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## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x+x^{2}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x+x^{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-2\right)(x+1)}{x(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{x^{2}-x-2}{x}=\frac{(-1)^{2}-(-1)-2}{-1}=\frac{1+1-2}{-1}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 10-2
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1-\cos 10 x}{e^{-x^{2}}-1}$
|
## Solution
Let's use the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& 1-\cos 10 x \sim \frac{(10 x)^{2}}{2}, \text{ as } x \rightarrow 0(10 x \rightarrow 0) \\
& \varepsilon^{x^{2}}-1 \sim x^{2}, \text{ as } x^{x} \rightarrow 0\left(x^{2} \rightarrow 0\right)
\end{aligned}
$$
We get:
$$
\lim _{x \rightarrow \overline{\overline{0}}} \frac{1-\cos 10 x}{e^{x^{2}}-1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow \overline{0}} \frac{\frac{100 x^{2}}{2}}{x^{2}}=\lim _{x \rightarrow \overline{0}} \frac{100}{2}=50
$$
## Problem Kuznetsov Limits 12-2
|
50
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-2 x}}{2 \arcsin x-\sin x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-2 x}}{2 \arcsin x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)}{2 \arcsin x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}(2 \arcsin x-\sin x)}=$
$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \arcsin x-\sin x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-2 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \arcsin x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}\right)=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
$e^{-2 x}-1 \sim-2 x$, as $x \rightarrow 0(-2 x \rightarrow 0)$
$\arcsin x \sim x$, as $x \rightarrow 0$
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{3 x}{x}-\lim _{x \rightarrow 0} \frac{-2 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 3-\lim _{x \rightarrow 0}-2}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{3+2}{2-1}=5$
## Problem Kuznetsov Limits 15-2
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{1+x \sin x-\cos 2 x}{\sin ^{2} x}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1+x \sin x-\cos 2 x}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{1+x \sin x-\left(1-2 \sin ^{2} x\right)}{\sin ^{2} x}= \\
& =\lim _{x \rightarrow 0} \frac{x \sin x+2 \sin ^{2} x}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{x \sin x}{\sin ^{2} x}+\lim _{x \rightarrow 0} \frac{2 \sin ^{2} x}{\sin ^{2} x}= \\
& =\lim _{x \rightarrow 0} \frac{x}{\sin x}+\lim _{x \rightarrow 0} 2=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$$
=\lim _{x \rightarrow 0} \frac{x}{x}+2=\lim _{x \rightarrow 0} 1+2=3
$$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{2+x}{3-x}\right)^{x}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{2+x}{3-x}\right)^{x}=\left(\frac{2+0}{3-0}\right)^{0}=\left(\frac{2}{3}\right)^{0}=1$
## Problem Kuznetsov Limits 18-2
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}}(\operatorname{tg} x)^{\operatorname{ctg} x}$
|
Solution
$\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{\cot x}=(\tan \frac{\pi}{4})^{\cot \frac{\pi}{4}}=1^{1}=1$
## Problem Kuznetsov Limits 20-2
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-4 ; 0 ; 4), B(-1 ; 6 ; 7), C(1 ; 10 ; 9)$
|
## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-1-(-4) ; 6-0 ; 7-4)=(3 ; 6 ; 3)$
$\overrightarrow{A C}=(1-(-4) ; 10-0 ; 9-4)=(5 ; 10 ; 5)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightarrow{A B}, \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$
$=\frac{3 \cdot 5+6 \cdot 10+3 \cdot 5}{\sqrt{3^{2}+6^{2}+3^{2}} \cdot \sqrt{5^{2}+10^{2}+5^{2}}}=$
$=\frac{15+60+15}{\sqrt{9+36+9} \cdot \sqrt{25+100+25}}=\frac{90}{\sqrt{54} \cdot \sqrt{150}}=1$
Thus, the cosine of the angle:
$\cos (\overrightarrow{A B,} \overrightarrow{A C})=1$
and consequently the angle
$\overrightarrow{A B}, \overrightarrow{A C}=0$
Problem Kuznetsov Analytic Geometry 4-30
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p-3 q$
$b=5 p+q$
$|p|=2$
$|q|=3$
$(\widehat{p, q})=\frac{\pi}{2}$
|
## 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:
$a \times b=(2 p-3 q) \times(5 p+q)=2 \cdot 5 \cdot p \times p+2 \cdot p \times q-3 \cdot 5 \cdot q \times p-3 \cdot q \times q=$ $=2 \cdot p \times q-15 \cdot q \times p=2 \cdot p \times q+15 \cdot p \times q=(2+15) \cdot p \times q=17 \cdot p \times q$
We compute the area:
$$
\begin{aligned}
& S=|a \times b|=|17 \cdot p \times q|=17 \cdot|p \times q|=17 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =17 \cdot 2 \cdot 3 \cdot \sin \frac{\pi}{2}=102 \cdot \sin \frac{\pi}{2}=102 \cdot 1=102
\end{aligned}
$$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 102.
## Problem Kuznetsov Analytic Geometry 5-30
|
102
|
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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(2 ; -4 ; -3) \)
\( A_{2}(5 ; -6 ; 0) \)
\( A_{3}(-1 ; 3 ; -3) \)
\( A_{4}(-10 ; -8 ; 7) \)
|
## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{5-2 ;-6-(-4) ; 0-(-3)\}=\{3 ;-2 ; 3\} \\
& A_{1} A_{3}=\{-1-2 ; 3-(-4) ;-3-(-3)\}=\{-3 ; 7 ; 0\} \\
& A_{1} \overrightarrow{A_{4}}=\{-10-2 ;-8-(-4) ; 7-(-3)\}=\{-12 ;-4 ; 10\}
\end{aligned}
$$
According to the geometric meaning of the mixed product, we have:
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot\left|\left(A_{1} A_{2}, A_{1} A_{3}, \vec{A}_{1} A_{4}\right)\right|
$$
We compute the mixed product:
$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
3 & -2 & 3 \\
-3 & 7 & 0 \\
-12 & -4 & 10
\end{array}\right|= \\
& =3 \cdot\left|\begin{array}{cc}
7 & 0 \\
-4 & 10
\end{array}\right|-(-2) \cdot\left|\begin{array}{cc}
-3 & 0 \\
-12 & 10
\end{array}\right|+3 \cdot\left|\begin{array}{cc}
-3 & 7 \\
-12 & -4
\end{array}\right|= \\
& =3 \cdot 70--2 \cdot(-30)+3 \cdot 96=210-60+288=438
\end{aligned}
$$
We obtain:
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|438|=73
$$
Since
$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{3} \cdot S_{A_{1} A_{2} A_{3}} \cdot h \Rightarrow h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}
$$
$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot\left|\overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}\right|
$$
We compute the vector product:
$$
\begin{aligned}
& \vec{A}_{1} A_{2} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
3 & -2 & 3 \\
-3 & 7 & 0
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-2 & 3 \\
7 & 0
\end{array}\right|-j\left|\begin{array}{cc}
3 & 3 \\
-3 & 0
\end{array}\right|+k \cdot\left|\begin{array}{cc}
3 & -2 \\
-3 & 7
\end{array}\right|= \\
& =i \cdot(-21)-j \cdot 9+k \cdot 15=\{-21 ;-9 ; 15\}
\end{aligned}
$$
We obtain:
$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-21)^{2}+(-9)^{2}+15^{2}}=\frac{\sqrt{747}}{2}
$$
Then
$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 73}{\frac{\sqrt{747}}{2}}=\frac{438}{\sqrt{747}}$
Volume of the tetrahedron: 73
Height: $\frac{438}{\sqrt{747}}$
## Problem Kuznetsov Analytic Geometry 7-30
|
73
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}}{\sqrt[4]{n^{5}+5}+\sqrt{n}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}}{\sqrt[4]{n^{5}+5}+\sqrt{n}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}\right)}{\frac{1}{n}\left(\sqrt[4]{n^{5}+5}+\sqrt{n}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{1-\frac{7}{n^{3}}}+\sqrt[3]{\frac{1}{n}+\frac{4}{n^{3}}}}{\sqrt[4]{n+\frac{5}{n^{4}}}+\sqrt{\frac{1}{n}}}=\left\{\frac{\sqrt[3]{1-0}+\sqrt[3]{0+0}}{\sqrt[4]{\infty+0}+\sqrt{0}}=\frac{1}{\infty}\right\}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 4-17
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-6 x^{2}+12 x-8}{x^{3}-3 x^{2}+4}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{x^{3}-6 x^{2}+12 x-8}{x^{3}-3 x^{2}+4}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}-4 x+4\right)}{(x-2)\left(x^{2}-x-2\right)}= \\
& =\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)^{2}}{(x-2)(x+1)}= \\
& =\lim _{x \rightarrow 2} \frac{x-2}{x+1}=\frac{2-2}{2+1}=0
\end{aligned}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-e^{\sin x}}{\tan x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-e^{\sin x}}{\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{\sin 2 x}-1\right)-\left(e^{\sin x}-1\right)}{\tan x}=$
$=\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-1}{\tan x}-\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{\tan x}=$
Using the substitution of equivalent infinitesimals:
$e^{\sin 2 x}-1 \sim \sin 2 x$, as $x \rightarrow 0(\sin 2 x \rightarrow 0)$
$e^{\sin x}-1 \sim \sin x$, as $x \rightarrow 0(\sin x \rightarrow 0)$
$\tan x \sim x$, as $x \rightarrow 0$
We get:
$=\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x}=$
Using the substitution of equivalent infinitesimals:
$\sin x \sim x$, as $x \rightarrow 0$
$\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
We get:
$=\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}=\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1=2-1=1$
## Problem Kuznetsov Limits 16-17
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-1 ; 2 ;-3), B(0 ; 1 ;-2), C(-3 ; 4 ;-5)$
|
## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-(-1) ; 1-2 ;-2-(-3))=(1 ;-1 ; 1)$
$\overrightarrow{A C}=(-3-(-1) ; 4-2 ;-5-(-3))=(-2 ; 2 ;-2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \cos (\overrightarrow{A B, \overrightarrow{A C}})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{1 \cdot(-2)+(-1) \cdot 2+1 \cdot(-2)}{\sqrt{1^{2}+(-1)^{2}+1^{2}} \cdot \sqrt{(-2)^{2}+2^{2}+(-2)^{2}}}= \\
& =\frac{-2-2-2}{\sqrt{1+1+1} \cdot \sqrt{4+4+4}}=\frac{-6}{\sqrt{3} \cdot \sqrt{12}}=-1
\end{aligned}
$$
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 4-24
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{2 x}{x^{2}+1}, x_{0}=1$
|
## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x}{x^{2}+1}\right)^{\prime}=\frac{2 x^{\prime}\left(x^{2}+1\right)-2 x\left(x^{2}+1\right)^{\prime}}{\left(x^{2}+1\right)^{2}}= \\
& =\frac{2 x^{2}+2-2 x \cdot 2 x}{\left(-2 x^{2}+2\right)^{2}}=\frac{2-2 x^{2}}{\left(x^{2}+1\right)^{2}}
\end{aligned}
$$
Then:
$$
y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2-2 \cdot 1^{2}}{\left(1^{2}+1\right)^{2}}=\frac{0}{2^{2}}=0
$$
Since the function $y^{\prime}{ }_{\text {at point }} x_{0}$ has a finite derivative, the equation of the tangent line is:
$$
\begin{aligned}
& y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right), \text { where } \\
& y_{0}^{\prime}=0 \\
& y_{0}=y\left(x_{0}\right)=\frac{2 \cdot 1}{1^{2}+1}=\frac{2}{2}=1
\end{aligned}
$$
We get:
$y-1=0 \cdot(x-(-2))$
$y=1$
That is, the equation of the tangent line is:
$y=1$
## Problem Kuznetsov Differentiation 3-23
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
\[
\begin{aligned}
& a=p-4 q \\
& b=3 p+q \\
& |p|=1 \\
& |q|=2 \\
& (\widehat{p, q})=\frac{\pi}{6}
\end{aligned}
\]
|
## 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:
$a \times b=(p-4 q) \times(3 p+q)=3 \cdot p \times p+p \times q-4 \cdot 3 \cdot q \times p-4 \cdot q \times q=$ $=p \times q-12 \cdot q \times p=p \times q+12 \cdot p \times q=(1+12) \cdot p \times q=13 \cdot p \times q$
We compute the area:
$$
\begin{aligned}
& S=|a \times b|=|13 \cdot p \times q|=13 \cdot|p \times q|=13 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =13 \cdot 1 \cdot 2 \cdot \sin \frac{\pi}{6}=26 \cdot \sin \frac{\pi}{6}=26 \cdot \frac{1}{2}=13
\end{aligned}
$$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 13.
## Problem Kuznetsov Analytic Geometry 5-9
|
13
|
Algebra
|
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 $a$?
$A(2; -5; 4)$
$a: 5x + 2y - z + 3 = 0$
$k = \frac{4}{3}$
|
## 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 $a$:
$a^{\prime}: 5 x+2 y-z+4=0$
Substitute the coordinates of point $A$ into the equation of $a^{\prime}$:
$5 \cdot 2+2 \cdot(-5)-4+4=0$
$10-10-4+4=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-9
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{2} \cos \frac{1}{\Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{2} \cos \frac{1}{\Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(2 \Delta x+\Delta x \cos \frac{1}{\Delta x}\right)=
\end{aligned}
$$
Since $\cos \frac{1}{\Delta x}$ 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 2-16
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2 ; 3 ; 1)$
$M_{2}(4 ; 1 ;-2)$
$M_{3}(6 ; 3 ; 7)$
$M_{0}(-5 ;-4 ; 8)$
|
## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-2 & y-3 & z-1 \\ 4-2 & 1-3 & -2-1 \\ 6-2 & 3-3 & 7-1\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-3 & z-1 \\
2 & -2 & -3 \\
4 & 0 & 6
\end{array}\right|=0 \\
& (x-2) \cdot\left|\begin{array}{cc}
-2 & -3 \\
0 & 6
\end{array}\right|-(y-3) \cdot\left|\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right|+(z-1) \cdot\left|\begin{array}{cc}
2 & -2 \\
4 & 0
\end{array}\right|=0 \\
& (x-2) \cdot(-12)-(y-3) \cdot 24+(z-1) \cdot 8=0 \\
& -12 x+24+(-24) y+72+8 z+(-8)=0 \\
& -12 x-24 y+8 z+88=0 \\
& -3 x-6 y+2 z+22=0
\end{aligned}
$$
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:
$$
d=\frac{|-3 \cdot(-5)-6 \cdot(-4)+2 \cdot 8+22|}{\sqrt{(-3)^{2}+(-6)^{2}+2^{2}}}=\frac{|15+24+16+22|}{\sqrt{9+36+4}}=\frac{77}{\sqrt{49}}=11
$$
## Problem Kuznetsov Analytic Geometry $8-29$
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}x^{2} \cos \left(\frac{4}{3 x}\right)+\frac{x^{2}}{2}, x \neq 0 \\ 0, x=0\end{array}\right.$
|
## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{x^{2} \cos \left(\frac{4}{3 x}\right)+\frac{x^{2}}{2}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{x^{2} \cos \left(\frac{4}{3 x}\right)+\frac{x^{2}}{2}}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(x \cos \left(\frac{4}{3 x}\right)+\frac{x}{2}\right)=
\end{aligned}
$$
Since $\cos \left(\frac{4}{3 x}\right)$ is bounded, then
$\Delta x \cdot \cos \left(\frac{4}{3 x}\right) \rightarrow 0 \quad$, as $\Delta x \rightarrow 0$
Then:
$=0+\frac{0}{2}=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation $2-8$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{2^{n+1}+3^{n+1}}{2^{n}+3^{n}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^{n+1}+3^{n+1}}{2^{n}+3^{n}}=\lim _{n \rightarrow \infty} \frac{2 \cdot 2^{n}+3 \cdot 3^{n}}{2^{n}+3^{n}}= \\
& =\lim _{n \rightarrow \infty} \frac{2 \cdot 2^{n}+2 \cdot 3^{n}+3^{n}}{2^{n}+3^{n}}=\lim _{n \rightarrow \infty} \frac{2\left(2^{n}+3^{n}\right)+3^{n}}{2^{n}+3^{n}}= \\
& =\lim _{n \rightarrow \infty}\left(2+\frac{3^{n}}{2^{n}+3^{n}}\right)=2+\lim _{n \rightarrow \infty} \frac{\frac{1}{3^{n}} 3^{n}}{\frac{1}{3^{n}}\left(2^{n}+3^{n}\right)}= \\
& =2+\lim _{n \rightarrow \infty} \frac{1}{\left(\frac{2}{3}\right)^{n}+1}=2+\frac{1}{0+1}=2+1=3
\end{aligned}
$$
## Problem Kuznetsov Limits 6-4
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1} \frac{\left(2 x^{2}-x-1\right)^{2}}{x^{3}+2 x^{2}-x-2}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\left(2 x^{2}-x-1\right)^{2}}{x^{3}+2 x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{(2 x+1)^{2}(x-1)^{2}}{\left(x^{2}+3 x+2\right)(x-1)}= \\
& =\lim _{x \rightarrow 1} \frac{(2 x+1)^{2}(x-1)}{x^{2}+3 x+2}=\frac{(2 \cdot 1+1)^{2}(1-1)}{1^{2}+3 \cdot 1+2}=\frac{3^{2} \cdot 0}{6}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 10-4
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{3 x}}{\sin 2 x-\sin x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{3 x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)}{\sin 2 x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\sin 2 x-\sin x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{5 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{5 x}-1 \sim 5 x$, as $x \rightarrow 0(5 x \rightarrow 0)$
$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
$\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{5 x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 5-\lim _{x \rightarrow 0} 3}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{5-3}{2-1}=2$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(2 ; 3 ; 2), B(-1 ;-3 ;-1), C(-3 ;-7 ;-3)$
|
## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-1-2 ;-3-3 ;-1-2)=(-3 ;-6 ;-3)$
$\overrightarrow{A C}=(-3-2 ;-7-3 ;-3-2)=(-5 ;-10 ;-5)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightarrow{A B}, \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$
$=\frac{(-3) \cdot(-5)+(-6) \cdot(-10)+(-3) \cdot(-5)}{\sqrt{(-3)^{2}+(-6)^{2}+(-3)^{2}} \cdot \sqrt{(-5)^{2}+(-10)^{2}+(-5)^{2}}}=$
$=\frac{15+60+15}{\sqrt{9+36+9} \cdot \sqrt{25+100+25}}=\frac{90}{\sqrt{54} \cdot \sqrt{150}}=1$
Thus, the cosine of the angle:
$\cos (\overrightarrow{A B,} \overrightarrow{A C})=1$
and consequently the angle
$\overrightarrow{A B}, \overrightarrow{A C}=0$
## Problem Kuznetsov Analytic Geometry 4-22
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p+4 q$
$b=q-p$
$|p|=2.5$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$
|
## 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:
$a \times b=(3 p+4 q) \times(q-p)=3 \cdot p \times q+3 \cdot(-1) \cdot p \times p+4 \cdot q \times q+4 \cdot(-1) \cdot q \times p=$ $=3 \cdot p \times q-4 \cdot q \times p=3 \cdot p \times q+4 \cdot p \times q=(3+4) \cdot p \times q=7 \cdot p \times q$
We compute the area:
$$
\begin{aligned}
& S=|a \times b|=|7 \cdot p \times q|=7 \cdot|p \times q|=7 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =7 \cdot 2.5 \cdot 2 \cdot \sin \frac{\pi}{2}=35 \cdot \sin \frac{\pi}{2}=35 \cdot 1=35
\end{aligned}
$$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 35.
## Problem Kuznetsov Analytic Geometry 5-22
|
35
|
Algebra
|
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 $a$?
$A(3 ; 5 ; 2)$
$a: 5x - 3y + z - 4 = 0$
$k = \frac{1}{2}$
|
## 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 $a$:
$a^{\prime}: 5 x-3 y+z-2=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}$:
$5 \cdot 3-3 \cdot 5+2-2=0$
$15-15+2-2=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-22
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
x^{2} e^{|x|} \sin \frac{1}{x^{2}}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} e^{|\Delta x|} \sin \frac{1}{\Delta x^{2}}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} e^{|\Delta x|} \cdot \Delta x \cdot \sin \frac{1}{\Delta x^{2}}= \\
& \quad \sin \frac{1}{\Delta x^{2}} \text { - is bounded, so }
\end{aligned}
$$
$\Delta x \sin \frac{1}{\Delta x^{2}} \rightarrow 0$, as $\Delta x \rightarrow 0$
Then:
$=e^{|0|} \cdot 0=1 \cdot 0=0$
Thus, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation $2-28$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\left((n+1)^{2}-(n-1)^{2}\right) \cdot\left((n+1)^{2}+(n-1)^{2}\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+2 n+1-n^{2}+2 n-1\right) \cdot\left(n^{2}+2 n+1+n^{2}-2 n+1\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n\left(2 n^{2}+2\right)}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}} 8 n\left(n^{2}+1\right)}{n^{3}\left((n+1)^{3}+(n-1)^{3}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{8\left(1+\frac{1}{n^{2}}\right)}{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}=\frac{8 \cdot 1}{1^{3}+1^{3}}=4
\end{aligned}
$$
## Problem Kuznetsov Limits 3-24
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}+2}-5 n^{2}}{n-\sqrt{n^{4}-n+1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}+2}-5 n^{2}}{n-\sqrt{n^{4}-n+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(\sqrt[3]{n^{2}+2}-5 n^{2}\right)}{\frac{1}{n^{2}}\left(n-\sqrt{n^{4}-n+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{\frac{1}{n^{4}}+\frac{2}{n^{6}}}-5}{\frac{1}{n}-\sqrt{1-\frac{1}{n^{3}}+\frac{1}{n^{4}}}}=\frac{\sqrt[3]{0+0}-5}{0-\sqrt{1-0+0}}=\frac{-5}{-1}=5
\end{aligned}
$$
## Problem Kuznetsov Limits 4-24
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2+4+6+\ldots+2 n}{1+3+5+\ldots+(2 n-1)}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2+4+6+\ldots+2 n}{1+3+5+\ldots+(2 n-1)}=\lim _{n \rightarrow \infty} \frac{\left(\frac{(2+2 n) n}{2}\right)}{\left(\frac{(1+(2 n-1)) n}{2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{(2+2 n) n}{(1+(2 n-1)) n}=\lim _{n \rightarrow \infty} \frac{2 n+2 n^{2}}{2 n^{2}}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{n}+1\right)=0+1=1
\end{aligned}
$$
## Problem Kuznetsov Limits 6-24
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sqrt[3]{8+3 x-x^{2}}-2}{\sqrt[3]{x^{2}+x^{3}}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt[3]{8+3 x-x^{2}}-2}{\sqrt[3]{x^{2}+x^{3}}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt[3]{8+3 x-x^{2}}-2\right)\left(\sqrt[3]{\left(8+3 x-x^{2}\right)^{2}}+2 \sqrt[3]{8+3 x-x^{2}}+4\right)}{\left(\sqrt[3]{x^{2}+x^{3}}\right)\left(\sqrt[3]{\left(8+3 x-x^{2}\right)^{2}}+2 \sqrt[3]{8+3 x-x^{2}}+4\right)}= \\
& =\lim _{x \rightarrow 0} \frac{8+3 x-x^{2}-8}{\sqrt[3]{x^{2}} \sqrt[3]{1+x}\left(\sqrt[3]{\left(8+3 x-x^{2}\right)^{2}}+2 \sqrt[3]{8+3 x-x^{2}}+4\right)}= \\
& =\lim _{x \rightarrow 0} \frac{x(3-x)}{\sqrt[3]{x^{2}} \sqrt[3]{1+x}\left(\sqrt[3]{\left(8+3 x-x^{2}\right)^{2}}+2 \sqrt[3]{8+3 x-x^{2}}+4\right)}= \\
& =\lim _{x \rightarrow 0} \frac{\sqrt[3]{x}(3-x)}{\sqrt[3]{1+x}\left(\sqrt[3]{\left(8+3 x-x^{2}\right)^{2}}+2 \sqrt[3]{8+3 x-x^{2}}+4\right)}= \\
& =\frac{\sqrt[3]{0}(3-0)}{\sqrt[3]{1+0}\left(\sqrt[3]{\left(8+3 \cdot 0-0^{2}\right)^{2}}+2 \sqrt[3]{8+3 \cdot 0-0^{2}}+4\right)}= \\
& =\frac{0}{1 \cdot\left(\sqrt[3]{8^{2}}+2 \sqrt[3]{8}+4\right)}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-24
This problem may have a different condition (possibly due to different editions or errors). See below for more details.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{1-\sin \left(\frac{x}{2}\right)}{\pi-x}$
|
## Solution
Substitution:
$$
\begin{aligned}
& x=y+\pi \Rightarrow y=x-\pi \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{1-\sin \left(\frac{x}{2}\right)}{\pi-x}=\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{y+\pi}{2}\right)}{\pi-(y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{y}{2}+\frac{\pi}{2}\right)}{-y}=\lim _{y \rightarrow 0} \frac{1-\cos \left(\frac{y}{2}\right)}{-y}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$1-\cos y \sim \frac{y^{2}}{2}$, as $y \rightarrow 0$
We get:
$=\lim _{y \rightarrow 0} \frac{\frac{y^{2}}{2}}{-y}=\lim _{y \rightarrow 0}-\frac{y}{2}=-\frac{0}{2}=0$
## Problem Kuznetsov Limits 13-24
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{e^{x}-e^{3 x}}{\sin 3 x-\tan 2 x}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{3 x}}{\sin 3 x-\tan 2 x}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{3 x}-1\right)}{\sin 3 x-\tan 2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}(\sin 3 x-\tan 2 x)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{3 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\sin 3 x-\tan 2 x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}-\lim _{x \rightarrow 0} \frac{\tan 2 x}{x}\right)=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$e^{x}-1 \sim x$, as $x \rightarrow 0$
$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
$\sin 3 x \sim 3 x$, as $3 x \rightarrow 0(3 x \rightarrow 0)$
$\tan 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightarrow 0} \frac{3 x}{x}-\lim _{x \rightarrow 0} \frac{2 x}{x}}=\frac{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0} 3}{\lim _{x \rightarrow 0} 3-\lim _{x \rightarrow 0} 2}=\frac{1-3}{3-2}=-2$
## Problem Kuznetsov Limits 15-24
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}= \\
& =\lim _{x \rightarrow \frac{\pi}{6}} \frac{2\left(\sin ^{2} x+\frac{1}{2} \cdot \sin x+\frac{1}{16}\right)-\frac{9}{8}}{2\left(\sin ^{2} x-\frac{3}{2} \sin x+\frac{9}{16}\right)-\frac{1}{8}}= \\
& =\lim _{x \rightarrow \frac{\pi}{6}} \frac{2\left(\sin x+\frac{1}{4}\right)^{2}-\frac{9}{8}}{2\left(\sin x-\frac{3}{4}\right)^{2}-\frac{1}{8}}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{2\left(\left(\sin x+\frac{1}{4}\right)^{2}-\frac{9}{16}\right)}{2\left(\left(\sin x-\frac{3}{4}\right)^{2}-\frac{1}{16}\right)}= \\
& =\lim _{x \rightarrow \frac{\pi}{6}} \frac{\left(\sin x+\frac{1}{4}-\frac{3}{4}\right)\left(\sin x+\frac{1}{4}+\frac{3}{4}\right)}{\left(\sin x-\frac{3}{4}-\frac{1}{4}\right)\left(\sin x-\frac{3}{4}+\frac{1}{4}\right)}= \\
& =\lim _{x \rightarrow \frac{\pi}{6}} \frac{\left(\sin x-\frac{1}{2}\right)(\sin x+1)}{(\sin x-1)\left(\sin x-\frac{1}{2}\right)}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin x+1}{\sin x-1}= \\
& =\frac{\sin \frac{\pi}{6}+1}{\sin \frac{\pi}{6}-1}=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3
\end{aligned}
$$
## Problem Kuznetsov Limits 16-24
|
-3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{\operatorname{arctg} 3 x}{x}\right)^{x+2}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\operatorname{arctg} 3 x}{x}\right)^{x+2}=\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} 3 x}{x}\right)^{\lim _{x \rightarrow 0} x+2}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} 3 x}{x}\right)^{0+2}=\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} 3 x}{x}\right)^{2}=
\end{aligned}
$$
$\operatorname{arctg} 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
We get:
$=\left(\lim _{x \rightarrow 0} \frac{3 x}{x}\right)^{2}=\left(\lim _{x \rightarrow 0} 3\right)^{2}=3^{2}=9$
## Problem Kuznetsov Limits 18-24
|
9
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \sqrt{\left(e^{\sin x}-1\right) \cos \left(\frac{1}{x}\right)+4 \cos x}$
|
## Solution
Since $\cos \left(\frac{1}{x}\right)_{\text {- is bounded, and }}$
$\lim _{x \rightarrow 0} e^{\sin x}-1=e^{\sin 0}-1=e^{0}-1=1-1=0$, then
$$
\left(e^{\sin x}-1\right) \cos \left(\frac{1}{x}\right) \rightarrow 0 \quad, \text { as } x \rightarrow 0
$$
Therefore:
$\lim _{x \rightarrow 0} \sqrt{\left(e^{\sin x}-1\right) \cos \left(\frac{1}{x}\right)+4 \cos x}=\sqrt{0+4 \cos 0}=\sqrt{4}=2$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the definite integral:
$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x
$$
|
## Solution
$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x=\int_{1 / 8}^{1} \frac{15}{(x+3) \sqrt{(x+3) x}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{x+3}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{1 \cdot x-(x+3) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{-3}{x^{2}} d x= \\
& =-\frac{3}{2 x \sqrt{(x+3) x}} d x \Rightarrow \frac{d x}{x \sqrt{(x+3) x}}=-\frac{2 d t}{3} \\
& x=\frac{1}{8} \Rightarrow t=\sqrt{\frac{\frac{1}{8}+3}{\frac{1}{8}}}=\sqrt{1+24}=5 \\
& x=1 \Rightarrow t=\sqrt{\frac{1+3}{1}}=\sqrt{4}=2
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{1 / 8}^{1} \frac{15}{\frac{x+3}{x} \cdot x \sqrt{(x+3) x}} d x=\int_{5}^{2} \frac{15}{t^{2}} \cdot\left(-\frac{2 d t}{3}\right)=\int_{5}^{2} \frac{-10 d t}{t^{2}}= \\
& =\left.\frac{10}{t}\right|_{5} ^{2}=\frac{10}{2}-\frac{10}{5}=5-2=3
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-15$ »
Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals
- Last modified: 21:56, 18 June 2009.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 11-16
## Material from PlusPi
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the definite integral:
$$
\int_{1 / 24}^{1 / 3} \frac{5 \sqrt{x+1}}{(x+1)^{2} \sqrt{x}} d x
$$
|
## Solution
$$
\int_{1 / 24}^{1 / 3} \frac{5 \sqrt{x+1}}{(x+1)^{2} \sqrt{x}} d x=\int_{1 / 24}^{1 / 3} \frac{5}{(x+1) \sqrt{(x+1) x}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{x+1}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+1}} \cdot \frac{1 \cdot x-(x+1) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{x+1}} \cdot \frac{-1}{x^{2}} d x= \\
& =-\frac{1}{2 x \sqrt{(x+1) x}} d x \Rightarrow \frac{d x}{x \sqrt{(x+1) x}}=-2 d t \\
& x=\frac{1}{24} \Rightarrow t=\sqrt{\frac{\frac{1}{24}+1}{\frac{1}{24}}}=\sqrt{1+24}=5 \\
& x=\frac{1}{3} \Rightarrow t=\sqrt{\frac{\frac{1}{3}+1}{\frac{1}{3}}}=\sqrt{1+3}=2
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{1 / 24}^{1 / 3} \frac{5}{\frac{x+1}{x} \cdot x \sqrt{(x+1) x}} d x=\int_{5}^{2} \frac{5}{t^{2}} \cdot(-2 d t)=\int_{5}^{2} \frac{-10 d t}{t^{2}}= \\
& =\left.\frac{10}{t}\right|_{5} ^{2}=\frac{10}{2}-\frac{10}{5}=5-2=3
\end{aligned}
$$
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}+11-22 »
Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals
- Last modified: 15:27, 26 June 2009.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 11-23
## Material from PlusPi
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the definite integral:
$$
\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x
$$
|
## Solution
## Problem 11.
## Calculate the definite integral:
$$
I=\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x
$$
## Perform a variable substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{x}{x-1}}, \text { hence } t^{2}=\frac{x}{x-1}, \text { that is } x=1+\frac{1}{t^{2}-1}, \text { and } \\
& d x=\frac{-2 t}{\left(t^{2}-1\right)^{2}} d t=-\frac{2 t d t}{\left(t^{2}-1\right)^{2}} . \\
& x=\frac{1}{t^{2}-1}+1=\frac{t^{2}}{t^{2}-1}
\end{aligned}
$$
Recalculate the limits of integration
$$
\begin{aligned}
& t_{1}=\sqrt{\frac{4 / 3}{4 / 3-1}}=2 \\
& t_{2}=\sqrt{\frac{16 / 15}{16 / 15-1}}=4
\end{aligned}
$$
## Take the resulting integral:
$$
I=\int_{4}^{2} \frac{4 t}{\frac{t^{4}}{\left(t^{2}-1\right)^{2}}} \cdot\left(-\frac{2 t}{\left(t^{2}-1\right)^{2}} d t\right)=8 \int_{2}^{4} \frac{1}{t^{2}} d t=-\left.\frac{8}{t}\right|_{2} ^{4}=-2+4=2
$$
Answer: $\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x=2$
\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_11-26»
Categories: In scan form | Kuznetsov's Integral Problems 11 | Integrals
Ukrainian Banner Network
- Last modified on this page: 11:23, June 8, 2009.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 11-27
## Material from Plusi
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+2)^{2}-(n-2)^{2}}{(n+3)^{2}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+2)^{2}-(n-2)^{2}}{(n+3)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{2}+4 n+4-n^{2}+4 n-4}{n^{2}+6 n+9}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n} 8 n}{\frac{1}{n}\left(n^{2}+6 n+9\right)}=\lim _{n \rightarrow \infty} \frac{8}{n+6+\frac{9}{n}}=\left\{\frac{8}{\infty}\right\}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 3-30
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+1}-\sqrt[3]{n^{3}+1}}{\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+1}-\sqrt[3]{n^{3}+1}}{\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+1}-\sqrt[3]{n^{3}+1}\right)}{\frac{1}{n}\left(\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}+\frac{1}{n^{2}}}-\sqrt[3]{1+\frac{1}{n^{3}}}}{\sqrt[4]{\frac{1}{n^{3}}+\frac{1}{n^{4}}}-\sqrt[5]{1+\frac{1}{n^{5}}}}=\frac{\sqrt{0+0}-\sqrt[3]{1+0}}{\sqrt[4]{0+0}-\sqrt[5]{1+0}}=\frac{-1}{-1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 4-30
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-3} \frac{x^{3}+7 x^{2}+15 x+9}{x^{3}+8 x^{2}+21 x+18}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-3} \frac{x^{3}+7 x^{2}+15 x+9}{x^{3}+8 x^{2}+21 x+18}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-3} \frac{(x+3)\left(x^{2}+4 x+3\right)}{(x+3)\left(x^{2}+5 x+6\right)}= \\
& =\lim _{x \rightarrow-3} \frac{x^{2}+4 x+3}{x^{2}+5 x+6}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-3} \frac{(x+3)(x+1)}{(x+3)(x+2)}= \\
& =\lim _{x \rightarrow-3} \frac{x+1}{x+2}=\frac{-3+1}{-3+2}=\frac{-2}{-1}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 10-30
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-8} \frac{10-x-6 \sqrt{1-x}}{2+\sqrt[3]{x}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-8} \frac{10-x-6 \sqrt{1-x}}{2+\sqrt[3]{x}}=\lim _{x \rightarrow-8} \frac{(10-x-6 \sqrt{1-x})(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})}= \\
& =\lim _{x \rightarrow-8} \frac{(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{100-20 x+x^{2}-36+36 x}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}= \\
& =\lim _{x \rightarrow-8} \frac{64+16 x+x^{2}}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{(8+x)^{2}}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=
\end{aligned}
$$
$$
\begin{aligned}
& =\lim _{x \rightarrow-8} \frac{\left((2+\sqrt[3]{x})\left(4-2 \sqrt[3]{x}+\sqrt[3]{x^{2}}\right)\right)^{2}}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{(2+\sqrt[3]{x})\left(4-2 \sqrt[3]{x}+\sqrt[3]{x^{2}}\right)^{2}}{10-x+6 \sqrt{1-x}}= \\
& =\frac{(2+\sqrt[3]{-8})\left(4-2 \sqrt[3]{-8}+\sqrt[3]{(-8)^{2}}\right)^{2}}{10-(-8)+6 \sqrt{1-(-8)}}=\frac{(2-2)\left(4-2 \cdot(-2)+(-2)^{2}\right)^{2}}{10+8+6 \sqrt{9}}= \\
& =\frac{0(4+4+4)^{2}}{10+8+6 \sqrt{9}}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-30
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-4 ; 2 ; 6)$
$M_{2}(2 ;-3 ; 0)$
$M_{3}(-10 ; 5 ; 8)$
$M_{0}(-12 ; 1 ; 8)$
|
## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-4) & y-2 & z-6 \\
2-(-4) & -3-2 & 0-6 \\
-10-(-4) & 5-2 & 8-6
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+4 & y-2 & z-6 \\
6 & -5 & -6 \\
-6 & 3 & 2
\end{array}\right|=0 \\
& (x+4) \cdot\left|\begin{array}{cc}
-5 & -6 \\
3 & 2
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
6 & -6 \\
-6 & 2
\end{array}\right|+(z-6) \cdot\left|\begin{array}{cc}
6 & -5 \\
-6 & 3
\end{array}\right|=0 \\
& (x+4) \cdot 8-(y-2) \cdot(-24)+(z-6) \cdot(-12)=0 \\
& 8 x+32+24 y-48-12 z+72=0 \\
& 8 x+24 y-12 z+56=0 \\
& 2 x+6 y-3 z+14=0
\end{aligned}
$$
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:
$$
d=\frac{|2 \cdot(-12)+6 \cdot 1-3 \cdot 8+14|}{\sqrt{2^{2}+6^{2}+(-3)^{2}}}=\frac{|-24+6-24+14|}{\sqrt{4+36+9}}=\frac{28}{\sqrt{49}}=4
$$
## Problem Kuznetsov Analytic Geometry $8-17$
|
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 $a?$
$A(2 ; 1 ; 2)$
$a: x-2 y+z+1=0$
$k=-2$
|
## 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 $a$:
$a^{\prime}: x-2 y+z-2=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}:$
$2-2 \cdot 1+2-2=0$
$2-2+2-2=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-2
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}-x-2}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-2\right)(x+1)}{x^{2}-x-2}= \\
& =\lim _{x \rightarrow-1}(x+1)=-1+1=0
\end{aligned}
$$
## Problem Kuznetsov Limits 10-9
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\cos 5 x-\cos 3 x}{\sin ^{2} x}$
|
## Solution
$\lim _{x \rightarrow \pi} \frac{\cos 5 x-\cos 3 x}{\sin ^{2} x}=\lim _{x \rightarrow \pi} \frac{-2 \sin \frac{5 x+3 x}{2} \sin \frac{5 x-3 x}{2}}{\sin ^{2} x}=$
$=\lim _{x \rightarrow \pi} \frac{-2 \sin 4 x \sin x}{\sin ^{2} x}=\lim _{x \rightarrow \pi} \frac{-2 \sin 4 x}{\sin x}=$
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarrow y \rightarrow 0$
We get:
$=\lim _{y \rightarrow 0} \frac{-2 \sin 4(y+\pi)}{\sin (y+\pi)}=\lim _{y \rightarrow 0} \frac{-2 \sin (4 y+4 \pi)}{-\sin y}=$
$=\lim _{y \rightarrow 0} \frac{2 \sin 4 y}{\sin y}=$
Using the substitution of equivalent infinitesimals:
$\sin 4 y \sim 4 y_{\text {, as }} y \rightarrow 0(4 y \rightarrow 0)$
$\sin y \sim y$, as $y \rightarrow 0$
$=\lim _{y \rightarrow 0} \frac{2 \cdot 4 y}{y}=\lim _{y \rightarrow 0} \frac{8}{1}=8$
## Problem Kuznetsov Limits 13-9
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}+1}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}+1}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}+1\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}{1+\frac{1}{n^{3}}}=\frac{1+1}{1}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 3-29
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n^{2}-\sqrt{n^{3}+1}}{\sqrt[3]{n^{6}+2}-n}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n^{2}-\sqrt{n^{3}+1}}{\sqrt[3]{n^{6}+2}-n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n^{2}-\sqrt{n^{3}+1}\right)}{\frac{1}{n^{2}}\left(\sqrt[3]{n^{6}+2}-n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{1-\sqrt{\frac{1}{n}+\frac{1}{n^{4}}}}{\sqrt[3]{1+\frac{2}{n^{6}}}-\frac{1}{n}}=\frac{1-\sqrt{0+0}}{\sqrt[3]{1+0}-0}=\frac{1}{1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 4-29
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)
$$
|
## Solution
$\lim _{n \rightarrow \infty} n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)=$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)\left(\sqrt{n^{4}+3}+\sqrt{n^{4}-2}\right)}{\sqrt{n^{4}+3}+\sqrt{n^{4}-2}}= \\
& =\lim _{n \rightarrow \infty} \frac{n\left(n^{4}+3-\left(n^{4}-2\right)\right)}{\sqrt{n^{4}+3}+\sqrt{n^{4}-2}}=\lim _{n \rightarrow \infty} \frac{5 n}{\sqrt{n^{4}+3}+\sqrt{n^{4}-2}}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}} 5 n}{\frac{1}{n^{2}}\left(\sqrt{n^{4}+3}+\sqrt{n^{4}-2}\right)}=\lim _{n \rightarrow \infty} \frac{\frac{5}{n}}{\sqrt{1+\frac{3}{n^{4}}}+\sqrt{1-\frac{2}{n^{4}}}}= \\
& =\frac{0}{\sqrt{1+0}+\sqrt{1-0}}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 5-29
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{x^{2}-16}}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{x^{2}-16}}=\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(x-4)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(\sqrt{x}-2)} \sqrt[3]{(\sqrt{x}+2)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt[3]{(\sqrt{x}-2)^{2}}}{\sqrt[3]{(\sqrt{x}+2)(x+4)}}=\frac{\sqrt[3]{(\sqrt{4}-2)^{2}}}{\sqrt[3]{(\sqrt{4}+2)(4+4)}}=
\end{aligned}
$$
$=\frac{\sqrt[3]{0^{2}}}{\sqrt[3]{(2+2) 8}}=0$
## Problem Kuznetsov Limits 11-29
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{x+\tan x^{2}}$
|
## 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
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=6 p-q$
$b=5 q+p$
$|p|=\frac{1}{2}$
$|q|=4$
$(\widehat{p, q})=\frac{5 \pi}{6}$
|
## 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:
$$
\begin{aligned}
& a \times b=(6 p-q) \times(5 q+p)=6 \cdot 5 \cdot p \times q+6 \cdot p \times p-5 \cdot q \times q-q \times p= \\
& =30 \cdot p \times q-q \times p=30 \cdot p \times q+p \times q=(30+1) \cdot p \times q=31 \cdot p \times q
\end{aligned}
$$
We compute the area:
$S=|a \times b|=|31 \cdot p \times q|=31 \cdot|p \times q|=31 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$
$=31 \cdot \frac{1}{2} \cdot 4 \cdot \sin \frac{5 \pi}{6}=62 \cdot \sin \frac{5 \pi}{6}=62 \cdot \frac{1}{2}=31$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 31.
## Problem Kuznetsov Analytical Geometry 5-28
|
31
|
Algebra
|
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 function:
$\lim _{x \rightarrow 0}\left(\frac{\tan 4 x}{x}\right)^{2+x}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{\tan 4 x}{x}\right)^{2+x}=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{\lim _{x \rightarrow 0} 2+x}=$
$=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{2+0}=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{2}=$
Using the substitution of equivalent infinitesimals:
$\tan 4 x \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$
We get:
$=\left(\lim _{x \rightarrow 0} \frac{4 x}{x}\right)^{2}=\left(\lim _{x \rightarrow 0} \frac{4}{1}\right)^{2}=4^{2}=16$
## Problem Kuznetsov Limits 18-8
|
16
|
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
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p-3 q$
$b=3 p+q$
$|p|=4$
$|q|=1$
$(\widehat{p, q})=\frac{\pi}{6}$
|
## 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:
$a \times b=(2 p-3 q) \times(3 p+q)=2 \cdot 3 \cdot p \times p+2 \cdot p \times q-3 \cdot 3 \cdot q \times p-3 \cdot q \times q=$ $=2 \cdot p \times q-9 \cdot q \times p=2 \cdot p \times q+9 \cdot p \times q=(2+9) \cdot p \times q=11 \cdot p \times q$
We compute the area:
$S=|a \times b|=|11 \cdot p \times q|=11 \cdot|p \times q|=11 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$
$=11 \cdot 4 \cdot 1 \cdot \sin \frac{\pi}{6}=44 \cdot \sin \frac{\pi}{6}=44 \cdot \frac{1}{2}=22$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 22.
## Problem Kuznetsov Analytic Geometry 5-16
|
22
|
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
|
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