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## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{x+\tan x^{2}}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{x+\tan x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{x+\tan x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}\left(x+\tan x^{2}\right)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(x+\tan x^{2}\right)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{\tan x^{2}}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{2 x}-1 \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{x}-1 \sim x$, as $x \rightarrow 0$
$\tan x^{2} \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{x^{2}}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0} x}=\frac{2-1}{1-0}=1$
## Problem Kuznetsov Limits 15-29
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{1}{x}\right)^{\frac{\ln (x+1)}{\ln (2-x)}}$
|
## Solution
$\lim _{x \rightarrow 1}\left(\frac{1}{x}\right)^{\frac{\ln (x+1)}{\ln (2-x)}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{1}{x}\right)}\right)^{\frac{\ln (x+1)}{\ln (2-x)}}=$
$=\lim _{x \rightarrow 1} e^{\frac{\ln (x+1)}{\ln (2-x)} \cdot \ln \left(\frac{1}{x}\right)}=\exp \left\{\lim _{x \rightarrow 1} \frac{\ln (x+1)}{\ln (2-x)} \cdot \ln \left(\frac{1}{x}\right)\right\}=$
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln ((y+1)+1)}{\ln (2-(y+1))} \cdot \ln \left(\frac{1}{y+1}\right)\right\}=$
$=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln (y+2)}{\ln (1-y)} \cdot \ln \left(\frac{1+y-y}{y+1}\right)\right\}=$
$=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln (y+2)}{\ln (1-y)} \cdot \ln \left(1-\frac{y}{y+1}\right)\right\}=$
Using the substitution of equivalent infinitesimals:
$\ln \left(1-\frac{y}{y+1}\right) \sim-\frac{y}{y+1}, \quad y \rightarrow 0\left(-\frac{y}{y+1} \rightarrow 0\right)$
$\ln (1-y) \sim-y$, as $y \rightarrow 0(-y \rightarrow 0)$
We get:
$=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln (y+2)}{-y} \cdot\left(-\frac{y}{y+1}\right)\right\}=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln (y+2)}{y+1}\right\}=$
$=\exp \left\{\frac{\ln (0+2)}{0+1}\right\}=\exp \left\{\frac{\ln 2}{1}\right\}=\exp \{\ln 2\}=e^{\ln 2}=2$
## Problem Kuznetsov Limits 19-29
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \sqrt{x\left(2+\sin \left(\frac{1}{x}\right)\right)+4 \cos x}$
|
## Solution
Since $2+\sin \left(\frac{1}{x}\right)_{\text { is bounded, then }}$
$x\left(2+\sin \left(\frac{1}{x}\right)\right) \rightarrow 0 \quad$, as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \sqrt{x\left(2+\sin \left(\frac{1}{x}\right)\right)+4 \cos x}=\sqrt{0+4 \cos 0}=\sqrt{4 \cdot 1}=2$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
1-\cos \left(x \sin \frac{1}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{1-\cos \left(\Delta x \sin \frac{1}{\Delta x}\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{1-\left(1-2 \sin ^{2}\left(\frac{1}{2} \cdot \Delta x \sin \frac{1}{\Delta x}\right)\right)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{2 \sin ^{2}\left(\frac{1}{2} \cdot \Delta x \sin \frac{1}{\Delta x}\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\sin \left(\frac{1}{2} \cdot \Delta x \sin \frac{1}{\Delta x}\right) \sim \frac{1}{2} \cdot \Delta x \sin \frac{1}{\Delta x}, \text { as } \Delta x \rightarrow 0\left(\frac{1}{2} \cdot \Delta x \sin \frac{1}{\Delta x} \rightarrow 0\right)
$$
We get:
$=\lim _{\Delta x \rightarrow 0} \frac{2 \cdot\left(\frac{1}{2} \cdot \Delta x \sin \frac{1}{\Delta x}\right)^{2}}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{1}{2} \cdot \Delta x \cdot \sin ^{2} \frac{1}{\Delta x}=$
Since $\sin ^{2} \frac{1}{\Delta x}$ is bounded, then
$\Delta x \cdot \sin ^{2} \frac{1}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$
Thus:
$=2 \cdot 0=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-31
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Condition of the problem
Are the vectors $a, b$ and $c$ coplanar?
$$
\begin{aligned}
& a=\{3 ; 3 ; 1\} \\
& b=\{1 ;-2 ; 1\} \\
& c=\{1 ; 1 ; 1\}
\end{aligned}
$$
|
## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
(a, b, c)=\left|\begin{array}{ccc}
3 & 3 & 1 \\
1 & -2 & 1 \\
1 & 1 & 1
\end{array}\right|=
$$
$=3 \cdot\left|\begin{array}{cc}-2 & 1 \\ 1 & 1\end{array}\right|-3 \cdot\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|+1 \cdot\left|\begin{array}{cc}1 & -2 \\ 1 & 1\end{array}\right|=$
$=3 \cdot(-3)-3 \cdot 0+1 \cdot 3=-9-0+3=-6$
Since $(a, b, c)=-6 \neq 0$, the vectors $a, b$ and $c$ are not coplanar.
## Problem Kuznetsov Analytic Geometry 6-5
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Find the angle between the planes:
$6 x+2 y-4 z+17=0$
$9 x+3 y-6 z-4=0$
|
## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{6 ; 2 ;-4\}$
$\overrightarrow{n_{2}}=\{9 ; 3 ;-6\}$
$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{6 \cdot 9+2 \cdot 3+(-4) \cdot(-6)}{\sqrt{6^{2}+2^{2}+(-4)^{2}} \cdot \sqrt{9^{2}+3^{2}+(-6)^{2}}}=$
$=\frac{54+6+24}{\sqrt{36+4+16} \cdot \sqrt{81+9+36}}=\frac{84}{\sqrt{56} \cdot \sqrt{126}}=\frac{84}{84}=1$
$\phi=\arccos 1=0$
## Problem Kuznetsov Analytic Geometry $10-5$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(1 ; \frac{1}{3} ;-2\right)$
$a: x-3 y+z+6=0$
$k=\frac{1}{3}$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-3 y+z+2=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}:$
$1-3 \cdot \frac{1}{3}-2+2=0$
$1-1-2+2=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry $12-5$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt[3]{27+x}-\sqrt[3]{27-x}}{\sqrt[3]{x^{2}}+\sqrt[5]{x}}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt[3]{27+x}-\sqrt[3]{27-x}}{\sqrt[3]{x^{2}}+\sqrt[3]{x}}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt[3]{27+x}-\sqrt[3]{27-x})\left(\sqrt[3]{(27+x)^{2}}+\sqrt[3]{27+x} \cdot \sqrt[3]{27-x}+\sqrt[3]{(27-x)^{2}}\right)}{\left(x^{\frac{2}{3}}+x^{\frac{1}{3}}\right)\left(\sqrt[3]{(27+x)^{2}}+\sqrt[3]{27+x} \cdot \sqrt[3]{27-x}+\sqrt[3]{(27-x)^{2}}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{27+x-(27-x)}{x^{\frac{1}{8}}\left(x^{\frac{2}{3}-\frac{1}{8}}+1\right)\left(\sqrt[3]{(27+x)^{2}}+\sqrt[3]{27+x} \cdot \sqrt[3]{27-x}+\sqrt[3]{(27-x)^{2}}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{2 x}{x^{\frac{1}{5}}\left(x^{\frac{7}{15}}+1\right)\left(\sqrt[3]{(27+x)^{2}}+\sqrt[3]{27+x} \cdot \sqrt[3]{27-x}+\sqrt[3]{(27-x)^{2}}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{2 x^{\frac{4}{8}}}{\left(x^{\frac{7}{15}}+1\right)\left(\sqrt[3]{(27+x)^{2}}+\sqrt[3]{27+x} \cdot \sqrt[3]{27-x}+\sqrt[3]{(27-x)^{2}}\right)}= \\
& =\frac{2 \cdot 0^{\frac{4}{8}}}{\left(0^{\frac{7}{15}}+1\right)\left(\sqrt[3]{(27+0)^{2}}+\sqrt[3]{27+0} \cdot \sqrt[3]{27-0}+\sqrt[3]{(27-0)^{2}}\right)}= \\
& =\frac{0}{1 \cdot\left(\sqrt[3]{27^{2}}+\sqrt[3]{27} \cdot \sqrt[3]{27}+\sqrt[3]{27^{2}}\right)}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-23
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\arcsin x}{x}\right)^{\frac{2}{x+5}}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{\arcsin x}{x}\right)^{\frac{2}{x+5}}=\left(\lim _{x \rightarrow 0} \frac{\arcsin x}{x}\right)^{\lim _{x \rightarrow 0} \frac{2}{x+5}}=$ $=\left(\lim _{x \rightarrow 0} \frac{\arcsin x}{x}\right)^{\frac{2}{0+5}}=\left(\lim _{x \rightarrow 0} \frac{\arcsin x}{x}\right)^{\frac{2}{5}}=$
Using the substitution of equivalent infinitesimals:
$\arcsin x \sim x$, as $x \rightarrow 0$
We get:
$=\left(\lim _{x \rightarrow 0} \frac{x}{x}\right)^{\frac{2}{5}}=\left(\lim _{x \rightarrow 0} 1\right)^{\frac{2}{5}}=1^{\frac{2}{5}}=1$
## Problem Kuznetsov Limits 18-23
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)^{\frac{1}{x^{2}}}$
|
$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 1}\left(\frac{(x-1)\left(x^{2}+x+1\right)}{x-1}\right)^{\frac{1}{x^{2}}}= \\
& =\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)^{\frac{1}{x^{2}}}=\left(1^{2}+1+1\right)^{\frac{1}{1^{2}}}=3^{1}=3
\end{aligned}
$$
The solution is as follows:
$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 1}\left(\frac{(x-1)\left(x^{2}+x+1\right)}{x-1}\right)^{\frac{1}{x^{2}}}= \\
& =\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)^{\frac{1}{x^{2}}}=\left(1^{2}+1+1\right)^{\frac{1}{1^{2}}}=3^{1}=3
\end{aligned}
$$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(-1 ; 2 ; 4) \)
\( A_{2}(-1 ;-2 ;-4) \)
\( A_{3}(3 ; 0 ;-1) \)
\( A_{4}(7 ;-3 ; 1) \)
|
## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-1-(-1) ;-2-2 ;-4-4\}=\{0 ;-4 ;-8\} \\
& \vec{A}_{1} A_{3}=\{3-(-1) ; 0-2 ;-1-4\}=\{4 ;-2 ;-5\} \\
& \overrightarrow{A_{1} A_{4}}=\{7-(-1) ;-3-2 ; 1-4\}=\{8 ;-5 ;-3\}
\end{aligned}
$$
According to the geometric meaning of the scalar triple product, we have:
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot\left|\left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)\right|
$$
We compute the scalar triple product:
$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
0 & -4 & -8 \\
4 & -2 & -5 \\
8 & -5 & -3
\end{array}\right|= \\
& =0 \cdot\left|\begin{array}{ll}
-2 & -5 \\
-5 & -3
\end{array}\right|-(-4) \cdot\left|\begin{array}{cc}
4 & -5 \\
8 & -3
\end{array}\right|+(-8) \cdot\left|\begin{array}{cc}
4 & -2 \\
8 & -5
\end{array}\right|= \\
& =0 \cdot(-19)--4 \cdot 28+-8 \cdot(-4)=0-(-112)+32=144
\end{aligned}
$$
We obtain:
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|144|=24
$$
Since
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{3} \cdot S_{A_{1} A_{2} A_{3}} \cdot h \Rightarrow h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}
$$
According to the geometric meaning of the vector product:
$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot\left|\overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}\right|$
We compute the vector product:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
0 & -4 & -8 \\
4 & -2 & -5
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-4 & -8 \\
-2 & -5
\end{array}\right|-j\left|\begin{array}{cc}
0 & -8 \\
4 & -5
\end{array}\right|+k \cdot\left|\begin{array}{cc}
0 & -4 \\
4 & -2
\end{array}\right|= \\
& =i \cdot 4-j \cdot 32+k \cdot 16=\{4 ;-32 ; 16\}
\end{aligned}
$$
We obtain:
$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{4^{2}+(-32)^{2}+16^{2}}=\frac{1}{2} \cdot \sqrt{1296}=\frac{36}{2}=18
$$
Then:
$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 24}{18}=4$
Volume of the tetrahedron: 24
Height: 4
## Problem Kuznetsov Analytic Geometry $7-25$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(2; -5; -1)$
$a: 5x + 2y - 3z - 9 = 0$
$k = \frac{1}{3}$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0$ and the coefficient $k$, the plane transitions to
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 5 x+2 y-3 z-3=0$
Substitute the coordinates of point $A$ into the equation of $a^{\prime}$:
$5 \cdot 2+2 \cdot(-5)-3 \cdot(-1)-3=0$
$10-10+3-3=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-25
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{2}+3 x+2\right)^{2}}{x^{3}+2 x^{2}-x-2}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\left(x^{2}+3 x+2\right)^{2}}{x^{3}+2 x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+2)^{2}(x+1)^{2}}{\left(x^{2}+x-2\right)(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{(x+2)^{2}(x+1)}{x^{2}+x-2}=\frac{(-1+2)^{2}(-1+1)}{(-1)^{2}+(-1)-2}=\frac{1 \cdot 0}{-2}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 10-3
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sqrt{x-1}}{\sqrt[3]{x^{2}-1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt{x-1}}{\sqrt[3]{x^{2}-1}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{\sqrt{x-1} \sqrt{x+1}}{\sqrt[3]{x^{2}-1} \sqrt{x+1}}= \\
& =\lim _{x \rightarrow 1} \frac{\sqrt{x^{2}-1}}{\sqrt[3]{x^{2}-1} \sqrt{x+1}}=\lim _{x \rightarrow 1} \frac{\sqrt[6]{\left(x^{2}-1\right)^{3}}}{\sqrt[6]{\left(x^{2}-1\right)^{2}} \sqrt{x+1}}= \\
& =\lim _{x \rightarrow 1} \frac{\sqrt[6]{x^{2}-1}}{\sqrt{x+1}}=\frac{\sqrt[6]{1^{2}-1}}{\sqrt{1+1}}=\frac{0}{2}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-3
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{3}+1}{\sin (x+1)}$
|
## Solution
Substitution:
$x=y-1 \Rightarrow y=x+1$
$x \rightarrow-1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}+1}{\sin (x+1)}=\lim _{y \rightarrow 0} \frac{(y-1)^{3}+1}{\sin ((y-1)+1)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{3}-3 y^{2}+3 y-1+1}{\sin y}=\lim _{y \rightarrow 0} \frac{y^{3}-3 y^{2}+3 y}{\sin y}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\sin y \sim y$, as $y \rightarrow 0$
We get:
$=\lim _{y \rightarrow 0} \frac{y^{3}-3 y^{2}+3 y}{y}=\lim _{y \rightarrow 0} y^{2}-3 y+3=0^{2}-3 \cdot 0+3=3$
## Problem Kuznetsov Limits $16-3$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x}\right)^{\frac{2}{x+2}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x}\right)^{\frac{2}{x+2}}=\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{\lim _{x \rightarrow 0} \frac{2}{x+2}}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{\frac{2}{0+2}}=\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{1}= \\
& =\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\sin 4 x \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$
We get:
$$
=\lim _{x \rightarrow 0} \frac{4 x}{x}=\lim _{x \rightarrow 0} \frac{4}{1}=4
$$
## Problem Kuznetsov Problem Kuznetsov Limits 18-3
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}}\left(\frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(x+\frac{\pi}{4}\right)}$
|
## Solution
$\lim _{x \rightarrow \frac{\pi}{4}}\left(\frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(x+\frac{\pi}{4}\right)}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{\lim _{x \rightarrow \frac{\pi}{4}} 1 /\left(x+\frac{\pi}{4}\right)}=$
$=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(\frac{\pi}{4}+\frac{\pi}{4}\right)}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(\frac{\pi}{2}\right)}=$
$=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (1-(1-\operatorname{tg} x))}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=$
Using the substitution of equivalent infinitesimals:
$\ln (1-(1-\operatorname{tg} x)) \sim-(1-\operatorname{tg} x)_{\text {, as }} x \rightarrow \frac{\pi}{4}(-(1-\operatorname{tg} x) \rightarrow 0)$
We get:
$$
\begin{aligned}
& =\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{-(1-\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\operatorname{tg} x-1}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}= \\
& =\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\operatorname{ctg} x}-\frac{\operatorname{ctg} x}{\operatorname{ctg} x}}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\operatorname{ctg} x}(1-\operatorname{ctg} x)}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}= \\
& =\left(\lim _{x \rightarrow \frac{\pi}{4}} \operatorname{tg} x\right)^{\frac{2}{\pi}}=\left(\operatorname{tg} \frac{\pi}{4}\right)^{\frac{2}{\pi}}=1^{\frac{2}{\pi}}=1
\end{aligned}
$$
Limits 20-3
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2 n-\sin n}{\sqrt{n}-\sqrt[3]{n^{3}-7}}$
|
## Solution
$$
\lim _{n \rightarrow \infty} \frac{2 n-\sin n}{\sqrt{n}-\sqrt[3]{n^{3}-7}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(2 n-\sin n)}{\frac{1}{n}\left(\sqrt{n}-\sqrt[3]{n^{3}-7}\right)}=
$$
$=\lim _{n \rightarrow x} \frac{2-\frac{\sin n}{n}}{\sqrt{\frac{1}{n}}-\sqrt[3]{1-\frac{7}{n^{3}}}}=$
Since $\sin n$ is bounded, then
$\sin n$
$\frac{\sin n}{n} \rightarrow 0$, as $n \rightarrow \infty$
Then:
$=\lim _{n \rightarrow \infty} \frac{2-0}{\sqrt{0}-\sqrt[3]{1-0}}=\lim _{n \rightarrow \infty} \frac{2}{0-1}=-2$
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-2 ; 4 ;-6), B(0 ; 2 ;-4), C(-6 ; 8 ;-10)$
|
## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-(-2) ; 2-4 ;-4-(-6))=(2 ;-2 ; 2)$
$\overrightarrow{A C}=(-6-(-2) ; 8-4 ;-10-(-6))=(-4 ; 4 ;-4)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightarrow{A B,} \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$
$=\frac{2 \cdot(-4)+(-2) \cdot 4+2 \cdot(-4)}{\sqrt{2^{2}+(-2)^{2}+2^{2}} \cdot \sqrt{(-4)^{2}+4^{2}+(-4)^{2}}}=$
$$
=\frac{-8-8-8}{\sqrt{4+4+4} \cdot \sqrt{16+16+16}}=\frac{-24}{2 \sqrt{3} \cdot 4 \sqrt{3}}=\frac{-24}{24}=-1
$$
Thus, the cosine of the angle:
$\cos (\overrightarrow{A B, A C})=-1$
and consequently the angle
$\widehat{A B,} \overrightarrow{A C}=\pi$
## Problem Kuznetsov Analytic Geometry 4-31
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 5 ;-7)$
$M_{2}(-3 ; 6 ; 3)$
$M_{3}(-2 ; 7 ; 3)$
$M_{0}(1 ;-1 ; 2)$
|
## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-1 & y-5 & z-(-7) \\ -3-1 & 6-5 & 3-(-7) \\ -2-1 & 7-5 & 3-(-7)\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-5 & z+7 \\
-4 & 1 & 10 \\
-3 & 2 & 10
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
1 & 10 \\
2 & 10
\end{array}\right|-(y-5) \cdot\left|\begin{array}{cc}
-4 & 10 \\
-3 & 10
\end{array}\right|+(z+7) \cdot\left|\begin{array}{ll}
-4 & 1 \\
-3 & 2
\end{array}\right|=0 \\
& (x-1) \cdot(-10)-(y-5) \cdot(-10)+(z+7) \cdot(-5)=0 \\
& -10 x+10+10 y-50-5 z-35=0 \\
& -10 x+10 y-5 z-75=0 \\
& 2 x-2 y+z+15=0
\end{aligned}
$$
The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$:
$d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$
Find:
$$
d=\frac{|2 \cdot 1-2 \cdot(-1)+2+15|}{\sqrt{2^{2}+(-2)^{2}+1^{2}}}=\frac{|2+2+2+15|}{\sqrt{4+4+1}}=\frac{21}{\sqrt{9}}=\frac{21}{3}=7
$$
## Problem Kuznetsov Analytic Geometry 8-31
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Find the distance from the point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$$
\begin{aligned}
& M_{1}(3 ; 10 ;-1) \\
& M_{2}(-2 ; 3 ;-5) \\
& M_{3}(-6 ; 0 ;-3) \\
& M_{0}(-6 ; 7 ;-10)
\end{aligned}
$$
|
## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-3 & y-10 & z-(-1) \\
-2-3 & 3-10 & -5-(-1) \\
-6-3 & 0-10 & -3-(-1)
\end{array}\right|=0
$$
Perform the transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-3 & y-10 & z-(-1) \\
-5 & -7 & -4 \\
-9 & -10 & -2
\end{array}\right|=0 \\
& (x-3) \cdot\left|\begin{array}{cc}
-7 & -4 \\
-10 & -2
\end{array}\right|-(y-10) \cdot\left|\begin{array}{cc}
-5 & -4 \\
-9 & -2
\end{array\right|+(z+1) \cdot\left|\begin{array}{cc}
-5 & -7 \\
-9 & -10
\end{array}\right|=0 \\
& (x-3) \cdot(-26)-(y-10) \cdot(-26)+(z+1) \cdot(-13)=0 \\
& -26 x+78+26 y-260-13 z-13=0 \\
& -26 x+26 y-13 z-195=0 \\
& -2 x+2 y-z-15=0
\end{aligned}
$$
The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$:
$d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$
$$
d=\frac{|-2 \cdot(-6)+2 \cdot 7-\cdot(-10)-15|}{\sqrt{(-2)^{2}+2^{2}+(-1)^{2}}}=\frac{|12+14+10-15|}{\sqrt{4+4+1}}=\frac{21}{\sqrt{9}}=7
$$
## Problem Kuznetsov Analytic Geometry $8-8$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\sqrt{1+\ln \left(1+3 x^{2} \cos \frac{2}{x}\right)}-1, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}-1-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\left(\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}-1\right) \cdot\left(\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1\right)}{\Delta x\left(\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1\right)}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)-1}{\Delta x\left(\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1\right)}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}{\Delta x\left(\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1\right)}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right) \sim 3 \Delta x^{2} \cos \frac{2}{\Delta x}_{\text {, as }} \Delta x \rightarrow 0\left(3 \Delta x^{2} \cos \frac{2}{\Delta x} \rightarrow 0\right)
$$
We get:
$$
=\lim _{\Delta x \rightarrow 0} \frac{3 \Delta x^{2} \cos \frac{2}{\Delta x}}{\Delta x\left(\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1\right)}=\lim _{\Delta x \rightarrow 0} \frac{3 \Delta x \cos \frac{2}{\Delta x}}{\sqrt{1+\ln \left(1+3 \Delta x^{2} \cos \frac{2}{\Delta x}\right)}+1}=
$$
Since $\cos \frac{2}{\Delta x}$ is bounded, then
$\Delta x \cdot \cos \frac{2}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$
Then:
$=\frac{3 \cdot 0}{\sqrt{1+\ln (1+3 \cdot 0)}+1}=\frac{0}{\sqrt{1}+1}=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-22
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=\frac{1+t^{3}}{t^{2}-1} \\
y=\frac{t}{t^{2}-1}
\end{array}
\right.
\]
$t_{0}=2$
|
## Solution
Since $t_{0}=2$, then
$x_{0}=\frac{1+2^{3}}{2^{2}-1}=\frac{9}{3}=3$
$y_{0}=\frac{2}{2^{2}-1}=\frac{2}{3}$
Let's find the derivatives:
$x_{t}^{\prime}=\left(\frac{1+t^{3}}{t^{2}-1}\right)^{\prime}=\frac{3 t^{2} \cdot\left(t^{2}-1\right)-\left(1+t^{3}\right) \cdot 2 t}{\left(t^{2}-1\right)^{2}}=\frac{3 t^{4}-3 t^{2}-2 t-2 t^{4}}{\left(t^{2}-1\right)^{2}}=$
$=\frac{t^{4}-3 t^{2}-2 t}{\left(t^{2}-1\right)^{2}}$
$y_{t}^{\prime}=\left(\frac{t}{t^{2}-1}\right)^{\prime}=\frac{1 \cdot\left(t^{2}-1\right)-t \cdot 2 t}{\left(t^{2}-1\right)^{2}}=\frac{t^{2}-1-2 t^{2}}{\left(t^{2}-1\right)^{2}}=$
$=-\frac{t^{2}+1}{\left(t^{2}-1\right)^{2}}$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{t^{2}+1}{\left(t^{2}-1\right)^{2}}\right) /\left(\frac{t^{4}-3 t^{2}-2 t}{\left(t^{2}-1\right)^{2}}\right)=-\frac{t^{2}+1}{t^{4}-3 t^{2}-2 t}$
Then:
$y_{0}^{\prime}=-\frac{2^{2}+1}{2^{4}-3 \cdot 2^{2}-2 \cdot 2}=-\frac{5}{16-12-4}=\infty$
Since $y_{0}^{\prime}=\infty$, the equation of the tangent line is:
$x=x_{0}$
$$
x=3
$$
The equation of the normal line:
$$
\begin{aligned}
& y-\frac{2}{3}=-\frac{1}{\infty} \cdot(x-3) \\
& y=-0 \cdot(x-3)+\frac{2}{3} \\
& y=\frac{2}{3}
\end{aligned}
$$
## Problem Kuznetsov Differentiation 17-22
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(1 ; 2 ; 2)$
$a: 3x - z + 5 = 0$
$k = -\frac{1}{5}$
|
## Solution
When transforming similarity with the center at the origin of the coordinate plane
$a: A x+B y+C z+D=0$ and the coefficient $k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-z-1=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}$:
$3 \cdot 1-2-1=0$
$3-2-1=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
Problem Kuznetsov Analytical Geometry 12-28
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\operatorname{arctg}\left(x^{3}-x^{\frac{3}{2}} \sin \frac{1}{3 x}\right), x \neq 0 ; \\ 0, x=0\end{array}\right.$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\operatorname{arctg}\left(\Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x}\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\operatorname{arctg}\left(\Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x}\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \operatorname{arctg}\left(\Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x}\right) \sim \Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x}, \text { as } \\
& \Delta x \rightarrow 0\left(\Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x} \rightarrow 0\right)
\end{aligned}
$$
We get:
$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{3}-\Delta x^{\frac{3}{2}} \sin \frac{1}{3 \Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(\Delta x^{2}-\Delta x^{\frac{1}{2}} \sin \frac{1}{3 \Delta x}\right)=$
Since $\sin \frac{1}{3 \Delta x}$ is bounded, then
$\Delta x^{\frac{1}{2}} \sin \frac{1}{3 \Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0\left(\Delta x^{\frac{1}{2}} \rightarrow 0\right)$
Then:
$=0^{2}-0=0$
Thus, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-9
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\sqrt{n(n+2)}-\sqrt{n^{2}-2 n+3}\right)
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\sqrt{n(n+2)}-\sqrt{n^{2}-2 n+3}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{n(n+2)}-\sqrt{n^{2}-2 n+3}\right)\left(\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}\right)}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}= \\
& =\lim _{n \rightarrow \infty} \frac{n(n+2)-\left(n^{2}-2 n+3\right)}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}=\lim _{n \rightarrow \infty} \frac{n^{2}+2 n-n^{2}+2 n-3}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n-3}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(4 n-3)}{\frac{1}{n\left(\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}\right)}}= \\
& =\lim _{n \rightarrow \infty} \frac{4-\frac{3}{n}}{\sqrt{1+\frac{2}{n}}+\sqrt{1-\frac{2}{n}+\frac{3}{n^{2}}}}=\frac{4-0}{\sqrt{1+0}+\sqrt{1-0+0}}=\frac{4}{2}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 5-8
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{x^{-2}-2 x+1}{2 x^{2}-x-1}$
|
## Solution
$\lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{2 x^{2}-x-1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{(x-1)^{2}}{(2 x+1)(x-1)}=$
$=\lim _{x \rightarrow 1} \frac{x-1}{2 x+1}=\frac{1-1}{2 \cdot 1+1}=\frac{0}{3}=0$
## Problem Kuznetsov Limits 10-8
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+x^{2}}-(1+x)}{x}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+x^{2}}-(1+x)}{x}=\left\{\frac{0}{0}\right\}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+x^{2}-(1+x)^{2}}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}=\lim _{x \rightarrow 0} \frac{1-2 x+x^{2}-\left(1^{2}+2 x+x^{2}\right)}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+x^{2}-1-2 x-x^{2}}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}=\lim _{x \rightarrow 0} \frac{-4 x^{2}}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{-4}{\sqrt{1-2 x+x^{2}}+(1+x)}=\frac{-4}{\sqrt{1-2 \cdot 0+0^{2}}+(1+0)}= \\
& =\frac{-4}{\sqrt{1}+1}=\frac{-4}{2}=-2
\end{aligned}
$$
## Problem Kuznetsov Limits 11-8
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 2 \pi} \frac{(x-2 \pi)^{2}}{\tan(\cos x-1)}
$$
|
## Solution
Substitution:
$x=y+2 \pi \Rightarrow y=x-2 \pi$
$x \rightarrow 2 \pi \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 2 \pi} \frac{(x-2 \pi)^{2}}{\operatorname{tg}(\cos x-1)}=\lim _{y \rightarrow 0} \frac{((y+2 \pi)-2 \pi)^{2}}{\operatorname{tg}(\cos (y+2 \pi)-1)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{2}}{\operatorname{tg}(\cos y-1)}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\operatorname{tg}(\cos y-1) \sim(\cos y-1)_{, \text {as }} y \rightarrow 0((\cos y-1) \rightarrow 0)$
We get:
$=\lim _{y \rightarrow 0} \frac{y^{2}}{\cos y-1}=$
Using the substitution of equivalent infinitesimals:
$1-\cos y \sim{\frac{y^{2}}{2}}_{, \text {as }} y \rightarrow 0$
We get:
$=\lim _{y \rightarrow 0} \frac{y^{2}}{-\frac{y^{2}}{2}}=\lim _{y \rightarrow 0} \frac{1}{-\frac{1}{2}}=-2$
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{4 x}-e^{-2 x}}{2 \operatorname{arctan} x-\sin x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{4 x}-e^{-2 x}}{2 \operatorname{arctg} x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)}{2 \operatorname{arctg} x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}(2 \operatorname{arctg} x-\sin x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \operatorname{arctg} x-\sin x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-2 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \operatorname{arctg} x}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{4 x}-1 \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$
$e^{-2 x}-1 \sim-2 x$, as $x \rightarrow 0(-2 x \rightarrow 0)$
$\operatorname{arctg} x \sim x$, as $x \rightarrow 0$
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{4 x}{x}-\lim _{x \rightarrow 0} \frac{-2 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 4-\lim _{x \rightarrow 0}-2}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{4+2}{2-1}=6$
## Problem Kuznetsov Limits 15-8
|
6
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}}{\sqrt[5]{n}-n}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}}{\sqrt[5]{n}-n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}\right)}{\frac{1}{n}(\sqrt[5]{n}-n)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{3}{n}-\frac{1}{n^{2}}}-\sqrt[3]{125+\frac{1}{n^{2}}}}{\sqrt[5]{\frac{1}{n^{4}}}-1}=\frac{\sqrt{0-0}-\sqrt[3]{125+0}}{\sqrt[5]{0}-1}=\frac{-5}{-1}=5
\end{aligned}
$$
## Problem Kuznetsov Limits 3-5(2)
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}}{\sqrt[3]{n}-n}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}}{\sqrt[3]{n}-n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}\right)}{\frac{1}{n}(\sqrt[3]{n}-n)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{3}{n}-\frac{1}{n^{2}}}-\sqrt[3]{125+\frac{1}{n^{2}}}}{\sqrt[3]{\frac{1}{n^{2}}}-1}=\frac{\sqrt{0-0}-\sqrt[3]{125+0}}{\sqrt[3]{0}-1}=\frac{-5}{-1}=5
\end{aligned}
$$
## Problem Kuznetsov Limits 4-5
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-3} \frac{\left(x^{2}+2 x-3\right)^{2}}{x^{3}+4 x^{2}+3 x}$
|
## Solution
$\lim _{x \rightarrow-3} \frac{\left(x^{2}+2 x-3\right)^{2}}{x^{3}+4 x^{2}+3 x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-3} \frac{(x+3)^{2}(x-1)^{2}}{x(x+1)(x+3)}=$
$=\lim _{x \rightarrow-3} \frac{(x+3)(x-1)^{2}}{x(x+1)}=\frac{(-3+3)(-3-1)^{2}}{-3(-3+1)}=\frac{0 \cdot(-4)^{2}}{6}=0$
## Problem Kuznetsov Limits 10-5
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\tan 2 x}-e^{-\sin 2 x}}{\sin x-1}$
|
## Solution
Substitution:
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\operatorname{tg} 2 x}-e^{-\sin 2 x}}{\sin x-1}=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2\left(y+\frac{\pi}{2}\right)}-e^{-\sin 2\left(y+\frac{\pi}{2}\right)}}{\sin \left(y+\frac{\pi}{2}\right)-1}$
$=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg}(2 y+\pi)}-e^{-\sin (2 y+\pi)}}{\cos y-1}=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}-e^{\sin 2 y}}{\cos y-1}=$
$=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}-e^{\frac{\sin 2 y \cos 2 y}{\cos 2 y}}}{\cos y-1}=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}-e^{\operatorname{tg} 2 y \cdot \cos 2 y}}{\cos y-1}=$
$=\lim _{y \rightarrow 0} \frac{-e^{\operatorname{tg} 2 y}\left(e^{\operatorname{tg} 2 y \cdot \cos 2 y-\operatorname{tg} 2 y}-1\right)}{\cos y-1}=$
$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}\left(e^{-\operatorname{tg} 2 y(1-\cos 2 y)}-1\right)}{1-\cos y}= \\
& =\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}\left(e^{-2 \operatorname{tg} 2 y \cdot \sin ^{2} y}-1\right)}{1-\cos y}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$e^{-2 \operatorname{tg} 2 y \cdot \sin ^{2} y}-1 \sim-2 \operatorname{tg} 2 y \cdot \sin ^{2} y$, as $y \rightarrow 0\left(-2 \operatorname{tg} 2 y \cdot \sin ^{2} y \rightarrow 0\right)$
$1-\cos y \sim \frac{y^{2}}{2}$, as $y \rightarrow 0$
We get:
$=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2 y}\left(-2 \operatorname{tg} 2 y \cdot \sin ^{2} y\right)}{\frac{y^{2}}{2}}=\lim _{y \rightarrow 0} \frac{-4 \operatorname{tg} 2 y \cdot \sin ^{2} y \cdot e^{\operatorname{tg} 2 y}}{y^{2}}=$
Using the substitution of equivalent infinitesimals:
$\sin y \sim y$, as $y \rightarrow 0$
$\operatorname{tg} 2 y \sim 2 y$, as $y \rightarrow 0(2 y \rightarrow 0)$
We get:
$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-4 \cdot 2 y \cdot y^{2} \cdot e^{\operatorname{tg} 2 y}}{y^{2}}=\lim _{y \rightarrow 0}-8 y e^{\operatorname{tg} 2 y}= \\
& =-8 \cdot 0 \cdot e^{\operatorname{tg} 2 \cdot 0}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 14-5
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}(\cos x)^{x+3}$
|
## Solution
$\lim _{x \rightarrow 0}(\cos x)^{x+3}=(\cos 0)^{0+3}=1^{3}=1$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 2}\left(\frac{\sin (3 \pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 2}\left(\frac{\sin (3 \pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}=\lim _{x \rightarrow 2}\left(\frac{\sin (2 \pi x+\pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}= \\
& =\lim _{x \rightarrow 2}\left(\frac{\sin 2 \pi x \cdot \cos \pi x+\cos 2 \pi x \cdot \sin \pi x}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}= \\
& =\lim _{x \rightarrow 2}\left(\frac{2 \sin \pi x \cdot \cos \pi x \cdot \cos \pi x+\cos 2 \pi x \cdot \sin \pi x}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}= \\
& =\lim _{x \rightarrow 2}\left(\frac{\sin \pi x\left(2 \cos ^{2} \pi x+\cos 2 \pi x\right)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}= \\
& =\lim _{x \rightarrow 2}\left(2 \cos ^{2} \pi x+\cos 2 \pi x\right)^{\sin ^{2}(x-2)}= \\
& =\left(2 \cos ^{2}(\pi \cdot 2)+\cos (2 \pi \cdot 2)\right)^{\sin ^{2}(2-2)}= \\
& =\left(2 \cdot 1^{2}+1\right)^{0^{2}}=3^{0}=1
\end{aligned}
$$
Problem Kuznetsov Limits 20-5
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}$
|
Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=\lim _{n \rightarrow \infty} \frac{4 n^{2}+4 n+1-n^{2}-2 n-1}{n^{2}+n+1}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 n^{2}+2 n\right)}{\frac{1}{n^{2}}\left(n^{2}+n+1\right)}=\lim _{n \rightarrow \infty} \frac{3+\frac{2}{n}}{1+\frac{1}{n}+\frac{1}{n^{2}}}=\frac{3+0}{1+0+0}=3
\end{aligned}
$$
## Problem Kuznetsov Limits 3-31
This problem may have a different condition (possibly due to different editions or errors).
For more details, see below
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{2+4+\ldots+2 n}{n+3}-n\right)$
|
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2+4+\ldots+2 n}{n+3}-n\right)=\lim _{n \rightarrow \infty}\left(\frac{\left(\frac{(2+2 n) n}{2}\right)}{n+3}-n\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{(1+n) n}{n+3}-n\right)=\lim _{n \rightarrow \infty}\left(\frac{(1+n) n}{n+3}-\frac{n(n+3)}{n+3}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{n+n^{2}-n^{2}-3 n}{n+3}=\lim _{n \rightarrow \infty} \frac{-2 n}{n+3}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-2) n}{\frac{1}{n}(n+3)}= \\
& =\lim _{n \rightarrow \infty} \frac{-2}{1+\frac{3}{n}}=-\frac{-2}{1+0}=-2
\end{aligned}
$$
## Problem Kuznetsov Limits 6-31
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 3} \frac{\sqrt{x+13}-2 \sqrt{x+1}}{\sqrt[3]{x^{2}-9}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{\sqrt{x+13}-2 \sqrt{x+1}}{\sqrt[3]{x^{2}-9}}=\lim _{x \rightarrow 3} \frac{(\sqrt{x+13}-2 \sqrt{x+1})(\sqrt{x+13}+2 \sqrt{x+1})}{\sqrt[3]{x^{2}-9}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{x+13-4(x+1)}{\sqrt[3]{x^{2}-9}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{-3 x+9}{\sqrt[3]{(x-3)(x+3)}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{-3(x-3)}{\sqrt[3]{x-3} \sqrt[3]{x+3}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{-3 \sqrt[3]{(x-3)^{2}}}{\sqrt[3]{x+3}(\sqrt{x+13}+2 \sqrt{x+1})}=\frac{-3 \sqrt[3]{(3-3)^{2}}}{\sqrt[3]{3+3}(\sqrt{3+13}+2 \sqrt{3+1})}= \\
& =\frac{-3 \sqrt[3]{0^{2}}}{\sqrt[3]{6}(\sqrt{16}+2 \sqrt{4})}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-31
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}$
|
## Solution
Let's use the substitution of equivalent infinitesimals:
$1-\cos x \sim \frac{x^{2}}{2}$, as $x \rightarrow 0$
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{2 x \cdot x}{\frac{x^{2}}{2}}=\lim _{x \rightarrow 0} \frac{2}{\frac{1}{2}}=4
$$
## Problem Kuznetsov Limits 12-31
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}$
|
## Solution
$\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan ^{2} 2 x}=\lim _{x \rightarrow \pi} \frac{-2 \sin \frac{3 x+x}{2} \sin \frac{3 x-x}{2}}{\tan ^{2} 2 x}=$
$=\lim _{x \rightarrow \pi} \frac{-2 \sin 2 x \sin x}{\tan ^{2} 2 x}=$
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarrow y \rightarrow 0$
We get:
$=\lim _{y \rightarrow 0} \frac{-2 \sin 2(y+\pi) \sin (y+\pi)}{\tan ^{2} 2(y+\pi)}=\lim _{y \rightarrow 0} \frac{2 \sin (2 y+2 \pi) \sin y}{\tan ^{2}(2 y+2 \pi)}=$
$=\lim _{y \rightarrow 0} \frac{2 \sin 2 y \sin y}{\tan ^{2} 2 y}=$
Using the substitution of equivalent infinitesimals:
$\sin y \sim y$, as $y \rightarrow 0$
$\sin 2 y \sim 2 y$, as $y \rightarrow 0$ (and $2 y \rightarrow 0$)
$\tan 2 y \sim 2 y$, as $y \rightarrow 0$ (and $2 y \rightarrow 0$)
We get:
$$
=\lim _{y \rightarrow 0} \frac{2 \cdot 2 y \cdot y}{(2 y)^{2}}=\lim _{y \rightarrow 0} \frac{4 y^{2}}{4 y^{2}}=\lim _{y \rightarrow 0} 1=1
$$
## Problem Kuznetsov Limits 13-31
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{2}+3 n-1}+\sqrt[3]{2 n^{2}+1}}{n+2 \sin n}$
|
## Solution
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{2}+3 n-1}+\sqrt[3]{2 n^{2}+1}}{n+2 \sin n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n^{2}+3 n-1}+\sqrt[3]{2 n^{2}+1}\right)}{\frac{1}{n}(n+2 \sin n)}=$
$=\lim _{n \rightarrow \infty} \frac{\sqrt{1+\frac{3}{n}-\frac{1}{n^{2}}}+\sqrt[3]{\frac{2}{n}+\frac{1}{n^{3}}}}{1+2 \frac{\sin n}{n}}=$
Since $\sin n$ is bounded, then
$\frac{\sin n}{n} \rightarrow 0$, as $n \rightarrow \infty$
Then:
$=\frac{\sqrt{1+0-0}+\sqrt[3]{0+0}}{1+2 \cdot 0}=\frac{\sqrt{1}+\sqrt[3]{0}}{1}=1$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(3, -6, 9), B(0, -3, 6), C(9, -12, 15)$
|
## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-3 ;-3-(-6) ; 6-9)=(-3 ; 3 ;-3)$
$\overrightarrow{A C}=(9-3 ;-12-(-6) ; 15-9)=(6 ;-6 ; 6)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \cos (\overrightarrow{A B, \overrightarrow{A C}})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{(-3) \cdot 6+3 \cdot(-6)+(-3) \cdot 6}{\sqrt{(-3)^{2}+3^{2}+(-3)^{2}} \cdot \sqrt{6^{2}+(-6)^{2}+6^{2}}}= \\
& =\frac{-18-18-18}{\sqrt{9+9+9} \cdot \sqrt{36+36+36}}=\frac{-54}{\sqrt{27} \cdot \sqrt{108}}=-\frac{54}{\sqrt{2916}}=-1
\end{aligned}
$$
Thus, the cosine of the angle:
$\cos (\overrightarrow{A B, A C})=-1$
and consequently the angle
$\widehat{A B,} \overrightarrow{A C}=\pi$
## Problem Kuznetsov Analytic Geometry 4-16
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; 5 ;-7) \)
\[
\begin{aligned}
& A_{2}(-3 ; 6 ; 3) \\
& A_{3}(-2 ; 7 ; 3) \\
& A_{4}(-4 ; 8 ;-12)
\end{aligned}
\]
|
## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-3-1 ; 6-5 ; 3-(-7)\}=\{-4 ; 1 ; 10\} \\
& A_{1} A_{3}=\{-2-1 ; 7-5 ; 3-(-7)\}=\{-3 ; 2 ; 10\} \\
& \overrightarrow{A_{1} A_{4}}=\{-4-1 ; 8-5 ;-12-(-7)\}=\{-5 ; 3 ;-5\}
\end{aligned}
$$
According to the geometric meaning of the scalar triple product, we have:
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot\left|\left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)\right|
$$
We compute the scalar triple product:
$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-4 & 1 & 10 \\
-3 & 2 & 10 \\
-5 & 3 & -5
\end{array}\right|= \\
& =-4 \cdot\left|\begin{array}{cc}
2 & 10 \\
3 & -5
\end{array}\right|-\left|\begin{array}{cc}
-3 & 10 \\
-5 & -5
\end{array}\right|+10 \cdot\left|\begin{array}{cc}
-3 & 2 \\
-5 & 3
\end{array}\right|= \\
& =-4 \cdot(-40)-65+10 \cdot 1=160-65+10=105
\end{aligned}
$$
We obtain:
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|105|=\frac{35}{2}=17.5
$$
Since
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{3} \cdot S_{A_{1} A_{2} A_{3}} \cdot h \Rightarrow h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}
$$
According to the geometric meaning of the vector product:
$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot\left|\overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}\right|
$$
We compute the vector product:
$$
\begin{aligned}
& \vec{A}_{1} A_{2} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-4 & 1 & 10 \\
-3 & 2 & 10
\end{array}\right|=i \cdot\left|\begin{array}{cc}
1 & 10 \\
2 & 10
\end{array}\right|-j\left|\begin{array}{cc}
-4 & 10 \\
-3 & 10
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-4 & 1 \\
-3 & 2
\end{array}\right|= \\
& =-10 \cdot i+10 \cdot j-5 \cdot k=\{-10 ; 10 ;-5\}
\end{aligned}
$$
We obtain:
$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-10)^{2}+10^{2}+(-5)^{2}}=\frac{15}{2}$
Then:
$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{35}{2}}{\frac{15}{2}}=\frac{3 \cdot 35}{15}=7$
Volume of the tetrahedron: 17.5
Height: 7
## Problem Kuznetsov Analytic Geometry 7-16
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=p-3q$
$b=p+2q$
$|p|=\frac{1}{5}$
$|q|=1$
$(\widehat{p, q})=\frac{\pi}{2}$
|
## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$$
S=|a \times b|
$$
We compute $a \times b$ using the properties of the vector product:
$$
\begin{aligned}
& a \times b=(p-3 q) \times(p+2 q)=p \times p+2 \cdot p \times q-3 \cdot q \times p-3 \cdot 2 \cdot q \times q= \\
& =2 \cdot p \times q-3 \cdot q \times p=2 \cdot p \times q+3 \cdot p \times q=(2+3) \cdot p \times q=5 \cdot p \times q
\end{aligned}
$$
We compute the area:
$$
\begin{aligned}
& S=|a \times b|=|5 \cdot p \times q|=5 \cdot|p \times q|=5 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =5 \cdot \frac{1}{5} \cdot 1 \cdot \sin \frac{\pi}{2}=\sin \frac{\pi}{2}=1
\end{aligned}
$$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 1.
## Problem Kuznetsov Analytic Geometry 5-3
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ; 5 ; 2\}$
$b=\{-1 ; 1 ;-1\}$
$c=\{1 ; 1 ; 1\}$
|
## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
(a, b, c)=\left|\begin{array}{ccc}
1 & 5 & 2 \\
-1 & 1 & -1 \\
1 & 1 & 1
\end{array}\right|=
$$
$=1 \cdot\left|\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right|-5 \cdot\left|\begin{array}{cc}-1 & -1 \\ 1 & 1\end{array}\right|+2 \cdot\left|\begin{array}{cc}-1 & 1 \\ 1 & 1\end{array}\right|=$
$=1 \cdot 2-5 \cdot 0+2 \cdot(-2)=2-0-4=-2$
Since $(a, b, c)=-2 \neq 0$, the vectors $a, b$ and $c$ are not coplanar.
## Problem Kuznetsov Analytic Geometry 6-3
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$$
\begin{aligned}
& M_{1}(-3 ;-1 ; 1) \\
& M_{2}(-9 ; 1 ;-2) \\
& M_{3}(3 ;-5 ; 4) \\
& M_{0}(-7 ; 0 ;-1)
\end{aligned}
$$
|
## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-3) & y-(-1) & z-1 \\
-9-(-3) & 1-(-1) & -2-1 \\
3-(-3) & -5-(-1) & 4-1
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+3 & y+1 & z-1 \\
-6 & 2 & -3 \\
6 & -4 & 3
\end{array}\right|=0 \\
& (x+3) \cdot\left|\begin{array}{cc}
2 & -3 \\
-4 & 3
\end{array}\right|-(y+1) \cdot\left|\begin{array}{cc}
-6 & -3 \\
6 & 3
\end{array}\right|+(z-1) \cdot\left|\begin{array}{cc}
-6 & 2 \\
6 & -4
\end{array}\right|=0 \\
& (x+3) \cdot(-6)-(y+1) \cdot 0+(z-1) \cdot 12=0 \\
& -6 x-18+12 z-12=0 \\
& -6 x+12 z-30=0 \\
& -x+2 z-5=0
\end{aligned}
$$
The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$:
$d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$
Find:
$$
d=\frac{|-1 \cdot(-7)+2 \cdot(-1)-5|}{\sqrt{(-1)^{2}+0^{2}+2^{2}}}=\frac{|7-2-5|}{\sqrt{1+0+4}}=\frac{0}{\sqrt{5}}=0
$$
## Problem Kuznetsov Analytic Geometry $8-3$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-1 ; 1 ; 1)$
$a: 3 x-y+2 z+4=0$
$k=\frac{1}{2}$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+2 z+2=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}$:
$3 \cdot(-1)-1+2 \cdot 1+2=0$
$-3-1+2+2=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-3
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\frac{\ln \left(1+2 x^{2}+x^{3}\right)}{x}, x \neq 0 ; \\ 0, x=0\end{array}\right.$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\frac{\ln \left(1+2 \Delta x^{2}+\Delta x^{3}\right)}{\Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\ln \left(1+2 \Delta x^{2}+\Delta x^{3}\right)}{\Delta x^{2}}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\ln \left(1+2 \Delta x^{2}+\Delta x^{3}\right) \sim 2 \Delta x^{2}+\Delta x^{3}, \text { as } \Delta x \rightarrow 0\left(2 \Delta x^{2}+\Delta x^{3} \rightarrow 0\right)
$$
We get:
$$
=\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{3}}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0}(2+\Delta x)=2+0=2
$$
Thus, $f^{\prime}(0)=2$
## Problem Kuznetsov Differentiation $2-29$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}$
|
Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+2}-\sqrt[3]{n^{3}+2}\right)}{\frac{1}{n}\left(\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}+\frac{2}{n^{2}}}-\sqrt[3]{1+\frac{2}{n^{3}}}}{\sqrt[7]{\frac{1}{n^{6}}+\frac{2}{n^{7}}}-\sqrt[5]{1+\frac{2}{n^{5}}}}=\frac{\sqrt{0+0}-\sqrt[3]{1+0}}{\sqrt[7]{0+0}-\sqrt[5]{1+0}}=\frac{-1}{-1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 4-25
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)\left(\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}\right)^{2}+\sqrt[3]{n^{2}\left(n^{6}+4\right)} \sqrt[3]{n^{8}-1}+\left(\sqrt[3]{n^{8}-1}\right)^{2}\right)}{\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}\right)^{2}+\sqrt[3]{n^{2}\left(n^{6}+4\right)} \sqrt[3]{n^{8}-1}+\left(\sqrt[3]{n^{8}-1}\right)^{2}}=
\end{aligned}
$$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(n^{2}\left(n^{6}+4\right)-\left(n^{8}-1\right)\right)}{\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(n^{8}+4 n^{2}-n^{8}+1\right)}{\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(4 n^{2}+1\right)}{\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{-\frac{1}{3}} n^{3}\left(4 n^{2}+1\right)}{n^{-\frac{16}{3}}\left(\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{-\frac{7}{3}}\left(4 n^{2}+1\right)}{\sqrt[3]{n^{-16} n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{-16} n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{n^{-16}\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n^{-\frac{1}{3}}+n^{-\frac{5}{3}}}{\sqrt[3]{\left(1+\frac{4}{n^{6}}\right)}+\sqrt[3]{\left(1+\frac{4}{n^{6}}\right)\left(1-\frac{1}{n^{8}}\right)}+\sqrt[3]{\left(1-\frac{1}{n^{8}}\right)^{2}}}= \\
& =\frac{4 \cdot 0+0}{\sqrt[3]{(1+0)^{2}}+\sqrt[3]{(1+0)(1-0)}+\sqrt[3]{(1-0)^{2}}}=\frac{0}{3}=0
\end{aligned}
$$
## Problem Kuznetsov Limits $5-25$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+3 x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+3 x^{2}-(1+x)^{2}}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}=\lim _{x \rightarrow 0} \frac{1-2 x+3 x^{2}-\left(1+2 x+x^{2}\right)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+3 x^{2}-1-2 x-x^{2}}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}=\lim _{x \rightarrow 0} \frac{-4 x+2 x^{2}}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}= \\
& =\lim _{x \rightarrow 0} \frac{x(2 x-4)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}=\lim _{x \rightarrow 0} \frac{\sqrt[3]{x^{2}}(2 x-4)}{\sqrt{1-2 x+3 x^{2}}+1+x}= \\
& =\frac{\sqrt[3]{0^{2}}(2 \cdot 0-4)}{\sqrt{1-2 \cdot 0+3 \cdot 0^{2}}+1+0}=\frac{0}{1+1}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-25
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sin ^{2} x-\tan^{2} x}{x^{4}}$
|
## Solution
We will use the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \sin x \sim x, \text { as } x \rightarrow 0 \\
& \operatorname{tg} x \sim x, \text { as } x \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin ^{2} x-\operatorname{tg}^{2} x}{x^{4}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{2} x}-\operatorname{tg}^{2} x}{x^{4}}= \\
& =\lim _{x \rightarrow 0} \frac{\operatorname{tg}^{2} x \cdot \cos ^{2} x-\operatorname{tg}^{2} x}{x^{4}}=\lim _{x \rightarrow 0} \frac{\operatorname{tg}^{2} x\left(\cos ^{2} x-1\right)}{x^{4}}=
\end{aligned}
$$
$$
=\lim _{x \rightarrow 0} \frac{x^{2}\left(-\sin ^{2} x\right)}{x^{4}}=\lim _{x \rightarrow 0} \frac{x^{2}\left(-x^{2}\right)}{x^{4}}=\lim _{x \rightarrow 0} \frac{-1}{1}=-1
$$
## Problem Kuznetsov Limits 12-25
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}=\left(e^{0}+0\right)^{\cos 0^{4}}= \\
& =(1+0)^{\cos 0}=1^{1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 18-25
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2}(\cos \pi x)^{\tan(x-2)}$
|
## Solution
$\lim _{x \rightarrow 2}(\cos \pi x)^{\operatorname{tg}(x-2)}=(\cos (\pi \cdot 2))^{\operatorname{tg}(2-2)}=(\cos 2 \pi)^{\operatorname{tg} 0}=1^{0}=1$
## Problem Kuznetsov Limits 20-25
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\ln \left(1-\sin \left(x^{3} \sin \frac{1}{x}\right)\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\ln \left(1-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\ln \left(1-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \ln \left(1-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)\right) \sim-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right), \text { as } \\
& \Delta x \rightarrow 0\left(-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right) \rightarrow 0\right)
\end{aligned}
$$
We get:
$=\lim _{\Delta x \rightarrow 0} \frac{-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)}{\Delta x}=$
Using the substitution of equivalent infinitesimals:
We get:
$$
=\lim _{\Delta x \rightarrow 0} \frac{-\Delta x^{3} \sin \frac{1}{\Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0}-\Delta x^{2} \sin \frac{1}{\Delta x}=
$$
Since $\sin \frac{1}{\Delta x}$ is bounded, then
$\Delta x^{2} \sin \frac{1}{\Delta x} \rightarrow{ }_{, \text {as }} \Delta x \rightarrow 0\left(\Delta^{2} x \rightarrow 0\right)$
Thus,
$=-0=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-4
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\arcsin (\sin t) \\
y=\arccos (\cos t)
\end{array}\right.
$$
|
## Solution
$x_{t}^{\prime}=(\arcsin (\sin t))^{\prime}=t^{\prime}=1$
$y_{t}^{\prime}=(\arccos (\cos t))^{\prime}=t^{\prime}=1$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1}{1}=1$
## Kuznetsov Differentiation 16-4
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in the Cartesian coordinate system.
$$
y=2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}, \frac{1}{4} \leq x \leq 1
$$
|
## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}\right)^{\prime} & =\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}} \cdot(1-2 x)= \\
& =\frac{1}{2 \sqrt{x} \sqrt{1-x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{2 x}{2 \sqrt{x} \sqrt{1-x}}= \\
& =\frac{1}{\sqrt{x} \sqrt{1-x}}-\frac{x}{\sqrt{x} \sqrt{1-x}}= \\
& =\frac{1-x}{\sqrt{x} \sqrt{1-x}}=\sqrt{\frac{1-x}{x}}
\end{aligned}
$$
Then, using the formula above, we get:
$$
\begin{aligned}
L & =\int_{1 / 4}^{1} \sqrt{1+\left(\sqrt{\frac{1-x}{x}}\right)^{2}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{1+\frac{1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{\frac{x+1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \frac{1}{\sqrt{x}} d x=\left.2 \sqrt{x}\right|_{1 / 4} ^{1}=2-\frac{2}{\sqrt{4}}=1
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�
\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-7$ »
Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals
- Last modified: 07:28, 26 May 2010.
- Content is available under CC-BY-SA 3.0.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=-\arccos \sqrt{x}+\sqrt{x-x^{2}}, 0 \leq x \leq \frac{1}{4}
$$
|
## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(-\arccos \sqrt{x}+\sqrt{x-x^{2}}\right)^{\prime} & =\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}} \cdot(1-2 x)= \\
& =\frac{1}{2 \sqrt{x} \sqrt{1-x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{2 x}{2 \sqrt{x} \sqrt{1-x}}= \\
& =\frac{1}{\sqrt{x} \sqrt{1-x}}-\frac{x}{\sqrt{x} \sqrt{1-x}}= \\
& =\frac{1-x}{\sqrt{x} \sqrt{1-x}}=\sqrt{\frac{1-x}{x}}
\end{aligned}
$$
Then, using the above formula, we get:
$$
\begin{aligned}
L & =\int_{1 / 4}^{1} \sqrt{1+\left(\sqrt{\frac{1-x}{x}}\right)^{2}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{1+\frac{1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{\frac{x+1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \frac{1}{\sqrt{x}} d x=\left.2 \sqrt{x}\right|_{1 / 4} ^{1}=2-\frac{2}{\sqrt{4}}=1
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{O} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D1} \% 82$ $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-14$ " Categories: Kuznetsov's Problem Book Integrals Problem 17| Integrals | Problems for checking
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- Last modified on this page: 07:46, 26 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 17-15
## Material from Plusi
## Contents
- 1 Problem Statement
- 2 Solution
- 3 Then
- 4 Using the substitution
- 5 Therefore
- 6 Using the substitution we get
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
x+\arcsin \left(x^{2} \sin \frac{6}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x+\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right) \sim \Delta x^{2} \sin \frac{6}{\Delta x} \text{ as } \Delta x \rightarrow 0 \left(\Delta x^{2} \sin \frac{6}{\Delta x} \rightarrow 0\right)$
We get:
$=\lim _{\Delta x \rightarrow 0} 1+\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} \sin \frac{6}{\Delta x}}{\Delta x}=1+\lim _{\Delta x \rightarrow 0} \Delta x \sin \frac{6}{\Delta x}=$
Since $\sin \frac{6}{\Delta x}$ is bounded, then
$\Delta x \sin \frac{6}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$
Thus:
$=1+0=1$
Therefore, $f^{\prime}(0)=1$
## Problem Kuznetsov Differentiation 2-11
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{\cos x-\cos 3 x}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\frac{\cos \Delta x-\cos (3 \Delta x)}{\Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\cos \Delta x-\cos (3 \Delta x)}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} \frac{1-2 \sin ^{2} \frac{\Delta x}{2}-\left(1-2 \sin ^{2}\left(\frac{3 \Delta x}{2}\right)\right)}{\Delta x^{2}}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2 \sin ^{2} \frac{\Delta x}{2}+2 \sin ^{2}\left(\frac{3 \Delta x}{2}\right)}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} \frac{-2 \sin ^{2} \frac{\Delta x}{2}}{\Delta x^{2}}+\lim _{\Delta x \rightarrow 0} \frac{2 \sin ^{2}\left(\frac{3 \Delta x}{2}\right)}{\Delta x^{2}}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\sin \frac{\Delta x}{2} \sim \frac{\Delta x}{2}$, as $\Delta x \rightarrow 0\left(\frac{\Delta x}{2} \rightarrow 0\right)$
$\sin \left(\frac{3 \Delta x}{2}\right) \sim \frac{3 \Delta x}{2}$, as $\Delta x \rightarrow 0\left(\frac{3 \Delta x}{2} \rightarrow 0\right)$
We get:
$$
=\lim _{\Delta x \rightarrow 0} \frac{-2\left(\frac{\Delta x}{2}\right)^{2}}{\Delta x^{2}}+\lim _{\Delta x \rightarrow 0} \frac{2\left(\frac{3 \Delta x}{2}\right)^{2}}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} \frac{-1}{2}+\lim _{\Delta x \rightarrow 0} \frac{9}{2}=-\frac{1}{2}+\frac{9}{2}=4
$$
Thus, $f^{\prime}(0)=4$
## Problem Kuznetsov Differentiation 2-30
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=\cos t+\sin t \\ y=\sin 2 t\end{array}\right.$
|
## Solution
$x_{t}^{\prime}=(\cos t+\sin t)^{\prime}=-\sin t+\cos t$
$y_{t}^{\prime}=(\sin 2 t)^{\prime}=2 \cos 2 t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 \cos 2 t}{-\sin t+\cos t}=2 \cdot \frac{\cos ^{2} t-\sin ^{2} t}{\cos t-\sin t}=2(\sin t+\cos t) \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=(2(\sin t+\cos t))^{\prime}=2(\cos t-\sin t)
\end{aligned}
$$
Then:
$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2(\cos t-\sin t)}{-\sin t+\cos t}=2$
## Problem Kuznetsov Differentiation 20-30
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}-3 n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}{1-\frac{3}{n^{2}}}=\frac{1^{3}+1^{3}}{1-0}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 3-28
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}=\lim _{n \rightarrow \infty} \frac{(n-1)!(n+n(n+1)(n+2))}{(n-1)!(1+n(n+1)(n+2))}= \\
& =\lim _{n \rightarrow \infty} \frac{n+n(n+1)(n+2)}{1+n(n+1)(n+2)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}(n+n(n+1)(n+2))}{\frac{1}{n^{3}}(1+n(n+1)(n+2))}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}+\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}{\frac{1}{n^{3}}+\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}=\frac{0+(1+0)(1+0)}{0+(1+0)(1+0)}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 6-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{1^{3}+3 \cdot 1^{2} \cdot x+3 \cdot 1 \cdot x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=$
$=\lim _{x \rightarrow 0} \frac{1+3 x+3 x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=\lim _{x \rightarrow 0} \frac{3 x^{2}+x^{3}}{x^{2}\left(1+x^{3}\right)}=$
$=\lim _{x \rightarrow 0} \frac{x^{2}(3+x)}{x^{2}\left(1+x^{3}\right)}=\lim _{x \rightarrow 0} \frac{3+x}{1+x^{3}}=\frac{3+0}{1+0^{3}}=\frac{3}{1}=3$
## Problem Kuznetsov Limits 10-28
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}=\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{x-6+8}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{x+2}{\sqrt[3]{(x+2)\left(x^{2}+2 x+4\right)}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{\sqrt[3]{(x+2)^{2}}}{\sqrt[3]{x^{2}+2 x+4}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\frac{\sqrt[3]{(-2+2)^{2}}}{\sqrt[3]{(-2)^{2}+2 \cdot(-2)+4}\left(\sqrt[3]{(-2-6)^{2}}-2 \sqrt[3]{-2-6}+4\right)}= \\
& =\frac{0}{\sqrt[3]{4-4+4}\left(\sqrt[3]{8^{2}}-2 \sqrt[3]{-8}+4\right)}=\frac{0}{\sqrt[3]{4}\left(2^{2}+2 \cdot 2+4\right)}=0
\end{aligned}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}$
|
## Solution
We will use the substitution of equivalent infinitesimals:
$\ln \left(1+x^{2}\right) \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{x^{2}}{1-\sqrt{x^{2}+1}}=$
$=\lim _{x \rightarrow 0} \frac{x^{2}\left(1+\sqrt{x^{2}+1}\right)}{\left(1-\sqrt{x^{2}+1}\right)\left(1+\sqrt{x^{2}+1}\right)}=$
$=\lim _{x \rightarrow 0} \frac{x^{2}\left(1+\sqrt{x^{2}+1}\right)}{1-\left(x^{2}+1\right)}=\lim _{x \rightarrow 0} \frac{x^{2}\left(1+\sqrt{x^{2}+1}\right)}{x^{2}}=$
$=\lim _{x \rightarrow 0}\left(1+\sqrt{x^{2}+1}\right)=\left(1+\sqrt{0^{2}+1}\right)=2$
## Problem Kuznetsov Limits 12-28
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{\sin 2 x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\sin 2 x-\sin x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{2 x}-1 \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{x}-1 \sim x$, as $x \rightarrow 0$
$\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$$
=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{2-1}{2-1}=1
$$
## Problem Kuznetsov Limits 15-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}=\left(6-\frac{5}{\cos 0}\right)^{\operatorname{tg}^{2} 0}= \\
& =\left(6-\frac{5}{1}\right)^{0^{2}}=1^{0}=1
\end{aligned}
$$
## Problem Kuznetsov Limits $18-28$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}$
|
## Solution
Substitution:
$$
\begin{aligned}
& x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2} \\
& x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0
\end{aligned}
$$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}=\lim _{y \rightarrow 0}\left(\sin \left(y+\frac{\pi}{2}\right)\right)^{\frac{18 \sin \left(y+\frac{\pi}{2}\right)}{\operatorname{ctg}\left(y+\frac{\pi}{2}\right)}}= \\
& =\lim _{y \rightarrow 0}(\cos y)^{\frac{18 \cos y}{-\operatorname{tg} y}}=\lim _{y \rightarrow 0}\left(e^{\ln (\cos y)}\right)^{-\frac{18 \cos y}{\operatorname{tg} y}}= \\
& =\lim _{y \rightarrow 0} e^{-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln (\cos y)}=\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln (\cos y)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln \left(1-2 \sin ^{2} \frac{y}{2}\right)\right\}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\ln \left(1-2 \sin ^{2} \frac{y}{2}\right) \sim-2 \sin ^{2} \frac{y}{2}$, as $\quad y \rightarrow 0\left(-2 \sin ^{2} \frac{y}{2} \rightarrow 0\right)$ $\operatorname{tg} y \sim y_{\text {, as }} y \rightarrow 0$
We get:
$=\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{y} \cdot\left(-2 \sin ^{2} \frac{y}{2}\right)\right\}=$
Using the substitution of equivalent infinitesimals:
$$
\sin \frac{y}{2} \sim \frac{y}{2}_{\text {as }} y \rightarrow 0\left(\frac{y}{2} \rightarrow 0\right)
$$
We get:
$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{36 \cos y}{y} \cdot\left(\frac{y}{2}\right)^{2}\right\}=\exp \left\{\lim _{y \rightarrow 0} 9 \cdot y \cdot \cos y\right\}= \\
& =\exp \{9 \cdot 0 \cdot \cos 0\}=\exp \{0\}=e^{0}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 19-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}
$$
|
## Solution
$\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}=(\sqrt[3]{1}+1-1)^{\sin \left(\frac{\pi \cdot 1}{4}\right)}=(1)^{\frac{\sqrt{2}}{2}}=1$
## Problem Kuznetsov Limits 20-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}
$$
|
## Solution
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sin t ; d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=\frac{1}{\sqrt{2}} \Rightarrow t=\arcsin \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 4} \frac{\cos t d t}{\sqrt{\left(1-\sin ^{2} t\right)^{3}}}=\int_{0}^{\pi / 4} \frac{\cos t d t}{\sqrt{\left(1-\sin ^{2} t\right)^{3}}}=\int_{0}^{\pi / 4} \frac{\cos t d t}{\cos ^{3} t}=\int_{0}^{\pi / 4} \frac{d t}{\cos ^{2} t}= \\
& =\left.\operatorname{tg} t\right|_{0} ^{\pi / 4}=\operatorname{tg} \frac{\pi}{4}-\operatorname{tg} 0=1
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-29$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals
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Created by Geeteatoo
## Problem Kuznetsov Integrals 12-30
## Material from PlusPi
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ;-1 ; 4\}$
$b=\{1 ; 0 ; 3\}$
$c=\{1 ;-3 ; 8\}$
|
## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}1 & -1 & 4 \\ 1 & 0 & 3 \\ 1 & -3 & 8\end{array}\right|=$
$=1 \cdot\left|\begin{array}{cc}0 & 3 \\ -3 & 8\end{array}\right|-(-1) \cdot\left|\begin{array}{ll}1 & 3 \\ 1 & 8\end{array}\right|+4 \cdot\left|\begin{array}{cc}1 & 0 \\ 1 & -3\end{array}\right|=$
$=1 \cdot 9+1 \cdot 5+4 \cdot(-3)=9+5-12=2$
Since $(a, b, c)=2 \neq 0$, the vectors $a, b$ and $c$ are not coplanar.
## Problem Kuznetsov Analytic Geometry 6-26
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}\right)}{\frac{1}{n}\left(\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}+\frac{3}{n^{2}}}-\sqrt[3]{8+\frac{3}{n^{3}}}}{\sqrt[4]{\frac{1}{n^{3}}+\frac{4}{n^{4}}}-\sqrt[5]{1+\frac{5}{n^{5}}}}=\frac{\sqrt{0+0}-\sqrt[3]{8+0}}{\sqrt[4]{0+0}-\sqrt[5]{1+0}}=\frac{-2}{-1}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 4-20
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}$
|
## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3) \cdot(2 n+2)!-(2 n+2)!}=$
$$
=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!((2 n+3)-1)}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!\cdot(2 n+2)}=
$$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(\frac{(2 n+1)!}{(2 n+2)!\cdot(2 n+2)}+\frac{(2 n+2)!}{(2 n+2)!\cdot(2 n+2)}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{(2 n+2) \cdot(2 n+2)}+\frac{1}{2 n+2}\right)=0+0=0
\end{aligned}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}$
|
## Solution
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}+2 x+1\right)}{x-2}=$
$=\lim _{x \rightarrow 2}\left(x^{2}+2 x+1\right)=2^{2}+2 \cdot 2+1=4+4+1=9$
## Problem Kuznetsov Limits 10-20
|
9
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)}{2 \sin x-\tan x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\frac{1}{x}(2 \sin x-\tan x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \sin x-\tan x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-5 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \sin x}{x}-\lim _{x \rightarrow 0} \frac{\tan x}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{2 x}-1 \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{-5 x}-1 \sim -5 x$, as $x \rightarrow 0(-5 x \rightarrow 0)$
$\sin x \sim x$, as $x \rightarrow 0$
$\tan x \sim x$, as $x \rightarrow 0$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{-5 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0}-5}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{2+5}{2-1}=7$
## Problem Kuznetsov Limits 15-20
|
7
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}$
|
## Solution
Since $\sin \left(\frac{1}{x}\right)_{\text {- is bounded as }} x \rightarrow 0$, then
$x \sin \left(\frac{1}{x}\right) \rightarrow 0 \quad$ as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}=\frac{2+\ln (e+0)}{\cos 0+\sin 0}=\frac{2+\ln e}{1+0}=\frac{2+1}{1+0}=3$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{2} ; \frac{1}{3} ; 1\right)$
$a: 2 x-3 y+3 z-2=0$
$k=1.5$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 2 x-3 y+3 z-3=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}$:
$2 \cdot \frac{1}{2}-3 \cdot \frac{1}{3}+3 \cdot 1-3=0$
$1-1+3-3=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
Problem Kuznetsov Analytical Geometry 12-6
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+6}-\sqrt{n^{2}-5}\right)}{\frac{1}{n}\left(\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}+\frac{6}{n^{2}}}-\sqrt{1-\frac{5}{n^{2}}}}{\sqrt[3]{1+\frac{3}{n^{3}}}+\sqrt[4]{\frac{1}{n}+\frac{1}{n^{4}}}}=\frac{\sqrt{0+0}-\sqrt{1-0}}{\sqrt[3]{1+0}+\sqrt[4]{0+0}}=\frac{-1}{1}=-1
\end{aligned}
$$
## Problem Kuznetsov Limits 4-27
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{7^{n}}\left(2^{n}+7^{n}\right)}{\frac{1}{7^{n}}\left(2^{n}-7^{n-1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\frac{2}{7}\right)^{n}+1}{\left(\frac{2}{7}\right)^{n}-\frac{1}{7}}=\frac{0+1}{0-\frac{1}{7}}=-7
\end{aligned}
$$
## Problem Kuznetsov Limits 6-27
|
-7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}=\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt[4]{x}-2)^{2}(\sqrt[4]{x}+2)^{2}}}= \\
& =\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{(\sqrt[4]{x}-2)^{\frac{2}{3}} \sqrt[3]{(\sqrt[4]{x}+2)^{2}}}=\lim _{x \rightarrow 16} \frac{(\sqrt[4]{x}-2)^{\frac{1}{3}}}{\sqrt[3]{(\sqrt[4]{x}+2)^{2}}}= \\
& =\frac{(\sqrt[4]{16}-2)^{\frac{1}{3}}}{\sqrt[3]{(\sqrt[4]{16}+2)^{2}}}=\frac{(2-2)^{\frac{1}{3}}}{\sqrt[3]{(2+2)^{2}}}=\frac{0}{\sqrt[3]{4^{2}}}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-27
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}=\left(\lim _{x \rightarrow 0} \tan\left(\frac{\pi}{4}-x\right)\right)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}= \\
& =\left(\tan\left(\frac{\pi}{4}-0\right)\right)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}=(1)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$e^{x}-1 \sim x$, as $x \rightarrow 0$
We get:
$=1^{\lim _{x \rightarrow 0} \frac{x}{x}}=1^{\lim _{x \rightarrow 0} 1}=1^{1}=1$
## Problem Kuznetsov Limits 18-27
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}$
|
## Solution
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}=\left(\cos \frac{\pi}{2}+1\right)^{\sin \frac{\pi}{2}}=(0+1)^{1}=1^{1}=1$
## Problem Kuznetsov Limits 20-27
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{9 x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{2} \cos \frac{1}{9 \Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{2} \cos \frac{1}{9 \Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(2 \Delta x+\Delta x \cos \frac{1}{9 \Delta x}\right)=
\end{aligned}
$$
Since $\cos \frac{1}{9 \Delta x}$ is bounded, then
$\Delta x \cdot \cos \frac{1}{9 \Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$
Then:
$=2 \cdot 0+0=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-14
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0):$
$f(x)=\left\{\begin{array}{c}\tan\left(x^{3}+x^{2} \sin \left(\frac{2}{x}\right)\right), x \neq 0 \\ 0, x=0\end{array}\right.$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\tan\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\tan\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \tan\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)\right) \sim \Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right), \text { as } \\
& \Delta x \rightarrow 0\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right) \rightarrow 0\right)
\end{aligned}
$$
We get:
$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \Delta x^{2}+\Delta x \sin \left(\frac{2}{\Delta x}\right)=$
Since $\sin \left(\frac{2}{\Delta x}\right)$ is bounded, then
$$
\Delta x \cdot \sin \left(\frac{2}{\Delta x}\right) \rightarrow 0 \quad \text { as } \Delta x \rightarrow 0
$$
Thus:
$=0^{2}+0=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-1
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\frac{4 x-x^{2}}{4}, x_{0}=2$
|
## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(\frac{4 x-x^{2}}{4}\right)^{\prime}=\frac{4-2 x}{4}=\frac{2-x}{2}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2-x_{0}}{2}=\frac{2-2}{2}=0$
Since $y^{\prime}\left(x_{0}\right)=0$, the equation of the normal line is:
$x=x_{0}$
$x=2$
Thus, the equation of the normal line is:
$x=2$
## Problem Kuznetsov Differentiation 3-1
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
3^{x^{2} \sin \frac{2}{x}}-1+2 x, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(3^{\Delta x^{2} \sin \frac{2}{\Delta x}}-1+2 x-0\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0}\left(\left(e^{\ln 3}\right)^{\Delta x^{2} \sin \frac{2}{\Delta x}}-1-2 \Delta x\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0}\left(e^{\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}}-1\right) / \Delta x-\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
e^{\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}}-1 \sim \Delta \ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}_{, \text {as }} \Delta x \rightarrow 0\left(\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x} \rightarrow 0\right)
$$
We get:
$$
\begin{aligned}
& =\lim _{\Delta x \rightarrow 0}\left(\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}\right) / \Delta x-\lim _{\Delta x \rightarrow 0} 2= \\
& =\lim _{\Delta x \rightarrow 0} \ln 3 \cdot \Delta x \sin \frac{2}{\Delta x}-2=
\end{aligned}
$$
Since $\sin \frac{2}{\Delta x}$ is bounded, then
$$
\Delta x \cdot \sin \frac{2}{\Delta x} \rightarrow 0, \text { as } \Delta x \rightarrow 0
$$
Thus
$=\ln 3 \cdot 0-2=-2$
Therefore, $f^{\prime}(0)=-2$
## Problem Kuznetsov Differentiation $2-21$
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}$
|
## Solution
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{3}+3 n^{2}+3 n+1-n^{3}+3 n^{2}-3 n+1}{n^{2}+2 n+1+n^{2}-2 n+1}=$
$=\lim _{n \rightarrow \infty} \frac{6 n^{2}+2}{2 n^{2}+2}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 n^{2}+1\right)}{\frac{1}{n^{2}}\left(n^{2}+1\right)}=\lim _{n \rightarrow \infty} \frac{3+\frac{1}{n^{2}}}{1+\frac{1}{n^{2}}}=\frac{3+0}{1+0}=3$
## Problem Kuznetsov Limits 3-26
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}\right)}{\frac{1}{n^{2}}(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{71}{n}}-\sqrt[3]{64+\frac{9}{n^{6}}}}{\left(\frac{1}{n}(n-\sqrt[3]{n})\right)\left(\frac{1}{n} \sqrt{11+n^{2}}\right)}=\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{71}{n}}-\sqrt[3]{64+\frac{9}{n^{6}}}}{\left(1-\sqrt[3]{\frac{1}{n^{2}}}\right) \sqrt{\frac{11}{n^{2}}+1}}= \\
& =\frac{\sqrt{0}-\sqrt[3]{64+0}}{(1-\sqrt[3]{0}) \sqrt{0+1}}=\frac{-4}{1}=-4
\end{aligned}
$$
## Problem Kuznetsov Limits 4-26
|
-4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}= \\
& =\{1-2=3-4=\ldots=(2 n-1)-2 n=-1\}= \\
& =\lim _{n \rightarrow \infty} \frac{-1 \cdot n}{\sqrt[3]{n^{3}+2 n+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-1) \cdot n}{\frac{1}{n} \sqrt[3]{n^{3}+2 n+2}}= \\
& =\lim _{n \rightarrow \infty} \frac{-1}{\sqrt[3]{1+\frac{2}{n^{2}}+\frac{2}{n^{3}}}}=\frac{-1}{\sqrt[3]{1+0+0}}=\frac{-1}{1}=-1
\end{aligned}
$$
## Problem Kuznetsov Limits 6-26
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)}{x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}\left(x+\sin x^{2}\right)}=
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(x+\sin x^{2}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-2 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{\sin x^{2}}{x}\right)=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$e^{x}-1 \sim x$, as $x \rightarrow 0$
$e^{-2 x}-1 \sim-2 x$, as $x \rightarrow 0(-2 x \rightarrow 0)$
$\sin x^{2} \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{-2 x}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{x^{2}}{x}}=\frac{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0}-2}{\lim _{x \rightarrow 0} 1+\lim _{x \rightarrow 0} x}=\frac{1+2}{1+0}=3$
## Problem Kuznetsov Limits 15-26
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\lim _{x \rightarrow 0} \frac{1}{x+6}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{0+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{6}}=$
Using the substitution of equivalent infinitesimals:
$\sin 5 x^{2} \sim 5 x^{2}$, as $x \rightarrow 0\left(5 x^{2} \rightarrow 0\right)$
$\sin x \sim x$, as $x \rightarrow 0(x \rightarrow 0)$
We get:
$=\left(\lim _{x \rightarrow 0} \frac{5 x^{2}}{x}\right)^{\frac{1}{6}}=\left(\lim _{x \rightarrow 0} 5 x\right)^{\frac{1}{6}}=0^{\frac{1}{6}}=0$
## Problem Kuznetsov Limits 18-26
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2; -1; 1)$
$a: x-2y+6z-10=0$
$k=\frac{3}{5}$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-2 y+6 z-6=0$
Substitute the coordinates of point $A$ into the equation of $a^{\prime}$:
$-2-2 \cdot(-1)+6 \cdot 1-6=0$
$-2+2+6-6=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-15
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}\right)}{\frac{1}{n^{2}}(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[5]{\frac{1}{n^{4}}}-\sqrt[3]{27+\frac{1}{n^{4}}}}{\left(\frac{1}{n}(n+\sqrt[4]{n})\right)\left(\frac{1}{n} \sqrt{9+n^{2}}\right)}=\lim _{n \rightarrow \infty} \frac{\sqrt[5]{\frac{1}{n^{4}}}-\sqrt[3]{27+\frac{1}{n^{4}}}}{\left(1+\sqrt[4]{\frac{1}{n^{3}}}\right) \sqrt{\frac{9}{n^{2}}+1}}= \\
& =\frac{\sqrt[5]{0}-\sqrt[3]{27+0}}{(1+\sqrt[4]{0}) \sqrt{0+1}}=\frac{-3}{1}=-3
\end{aligned}
$$
## Problem Kuznetsov Limits 4-6
|
-3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot \frac{(1+(2 n-1)) n}{2}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot n^{2}=\lim _{n \rightarrow \infty} \frac{1}{\frac{(1+n) n}{2}} \cdot n^{2}= \\
& =\lim _{n \rightarrow \infty} \frac{2 n}{1+n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} 2 n}{\frac{1}{n}(1+n)}=\lim _{n \rightarrow \infty} \frac{2}{\frac{1}{n}+1}= \\
& =\frac{2}{0+1}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 6-6
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}$
|
Solution
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)^{2}}{\left(x^{3}-x^{2}+x+1\right)(x+1)}=$
$=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)}{x^{3}-x^{2}+x+1}=\frac{\left((-1)^{2}-(-1)-1\right)^{2}(-1+1)}{(-1)^{3}-(-1)^{2}+(-1)+1}=$
$=\frac{(1+1-1)^{2} \cdot 0}{-1-1-1+1}=\frac{0}{-2}=0$
## Problem Kuznetsov Limits 10-6
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)}{\operatorname{arctg} x-x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}\left(\operatorname{arctg} x-x^{2}\right)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(\operatorname{arctg} x-x^{2}\right)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} x}{x}-\lim _{x \rightarrow 0} \frac{x^{2}}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{2 x}-1 \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
$\operatorname{arctg} x \sim x$, as $x \rightarrow 0$
We get:
$$
=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{x^{2}}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 3}{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0} x}=\frac{2-3}{1-0}=-1
$$
## Problem Kuznetsov Limits 15-6
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}=\lim _{x \rightarrow 0} \frac{\left(e^{\alpha x}-1\right)-\left(e^{\beta x}-1\right)}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
$=\lim _{x \rightarrow 0} \frac{e^{\alpha x}-1}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{e^{\beta x}-1}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
Using the substitution of equivalent infinitesimals:
$e^{\alpha x}-1 \sim \alpha x$, as $x \rightarrow 0(\alpha x \rightarrow 0)$
$e^{\beta x}-1 \sim \beta x$, as $x \rightarrow 0(\beta x \rightarrow 0)$
$\sin \frac{x(\alpha-\beta)}{2} \sim \frac{x(\alpha-\beta)}{2}, \quad x \rightarrow 0\left(\frac{x(\alpha-\beta)}{2} \rightarrow 0\right)$
We get:
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\alpha x}{2 \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{\beta x}{2 \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\alpha}{(\alpha-\beta) \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{\beta}{(\alpha-\beta) \cos \frac{x(\alpha+\beta)}{2}}= \\
& =\frac{\alpha}{(\alpha-\beta) \cos \frac{0 \cdot(\alpha+\beta)}{2}}-\frac{\beta}{(\alpha-\beta) \cos \frac{0 \cdot(\alpha+\beta)}{2}}= \\
& =\frac{\alpha}{(\alpha-\beta) \cos 0}-\frac{\alpha}{(\alpha-\beta) \cos 0}=\frac{\alpha}{(\alpha-\beta) \cdot 1}-\frac{\beta}{(\alpha-\beta) \cdot 1}= \\
& =\frac{\alpha-\beta}{\alpha-\beta}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 16-6
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}=\left(\frac{0^{2}+4}{0+2}\right)^{0^{2}+3}=$
$=\left(\frac{4}{2}\right)^{3}=2^{3}=8$
## Problem Kuznetsov Limits 18-6
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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