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## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{x+\tan x^{2}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{x+\tan x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{x+\tan x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}\left(x+\tan x^{2}\right)}=$
$=\frac{\lim _{... | 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... | 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Condition of the problem
Are the vectors $a, b$ and $c$ coplanar?
$$
\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|=
$$
$=3 \cdot\left|\begin{array}{cc}-2 &... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Find the angle between the planes:
$6 x+2 y-4 z+17=0$
$9 x+3 y-6 z-4=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{6 ; 2 ;-4\}$
$\overrightarrow{n_{2}}=\{9 ; 3 ;-6\}$
$
$a: x-3 y+z+6=0$
$k=\frac{1}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-3 y+z+2=0$
Substitute the coordinates of point $A$ into the equatio... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt[3]{27+x}-\sqrt[3]{27-x}}{\sqrt[3]{x^{2}}+\sqrt[5]{x}}$ | ## Solution
$$
\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}... | 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)^{\f... | 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{aligne... | 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 mean... | 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 equati... | 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
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\left(x^{2}+3 x+2\right)^{2}}{x^{3}+2 x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+2)^{2}(x+1)^{2}}{\left(x^{2}+x-2\right)(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{(x+2)^{2}(x+1)}{x^{2}+x-2}=\frac{(-1+2)^{2}(-1+1)}{(-1)^{2}+(-1)... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sqrt{x-1}}{\sqrt[3]{x^{2}-1}}$ | ## Solution
$$
\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{\s... | 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:
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}+1}{\sin (x+1)}=\lim _{y \rightarrow 0} \frac{(y-1)^{3}+1}{\sin ((y-1)+1)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{3}-3 y^{2}+3 y-1+1}{\sin y}=\lim _{y \... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\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 | 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}... | 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 $\s... | -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 (\ov... | -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:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-5 & z+7 \\
-4 & 1 & 10 \\
-3 & ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Find the distance from the point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$$
\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) \\
-... | 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} ... | 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... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(1 ; 2 ; 2)$
$a: 3x - z + 5 = 0$
$k = -\frac{1}{5}$ | ## Solution
When transforming similarity with the center at the origin of the coordinate plane
$a: A x+B y+C z+D=0$ and the coefficient $k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-z-1=0$
Substitute the coordinates of point $A$ into the eq... | 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\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(... | 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
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+x^{2}}-(1+x)}{x}=\left\{\frac{0}{0}\right\}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+x^{2... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\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)}= \\
& =\... | -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}... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}}{\sqrt[5]{n}-n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}}{\sqrt[5]{n}-n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}\right)}{\frac{1}{n}(\sqrt[5]{n}-n)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{3}{n}-\frac{1}{n^{2}}}-... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\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}}}-... | 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 Kuz... | 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... | 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
$$
\begin{aligned}
& \lim _{x \rightarrow 2}\left(\frac{\sin (3 \pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}=\lim _{x \rightarrow 2}\left(\frac{\sin (2 \pi x+\pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}= \\
& =\lim _{x \rightarrow 2}\left(\frac{\sin 2 \pi x \cdot \cos \pi x+\cos 2 \pi x \cdot \sin \pi x}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}$ | Solution
$$
\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 \inf... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{2+4+\ldots+2 n}{n+3}-n\right)$ | $$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2+4+\ldots+2 n}{n+3}-n\right)=\lim _{n \rightarrow \infty}\left(\frac{\left(\frac{(2+2 n) n}{2}\right)}{n+3}-n\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{(1+n) n}{n+3}-n\right)=\lim _{n \rightarrow \infty}\left(\frac{(1+n) n}{n+3}-\frac{n(n+3)}{n+3... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\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 ... | 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:
$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{2 x \cdot x}{\frac{x^{2}}{2}}=\lim... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}$ | ## Solution
$\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan ^{2} 2 x}=\lim _{x \rightarrow \pi} \frac{-2 \sin \frac{3 x+x}{2} \sin \frac{3 x-x}{2}}{\tan ^{2} 2 x}=$
$=\lim _{x \rightarrow \pi} \frac{-2 \sin 2 x \sin x}{\tan ^{2} 2 x}=$
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{2}+3 n-1}+\sqrt[3]{2 n^{2}+1}}{n+2 \sin n}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{2}+3 n-1}+\sqrt[3]{2 n^{2}+1}}{n+2 \sin n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n^{2}+3 n-1}+\sqrt[3]{2 n^{2}+1}\right)}{\frac{1}{n}(n+2 \sin n)}=$
$=\lim _{n \rightarrow \infty} \frac{\sqrt{1+\frac{3}{n}-\frac{1}{n^{2}}}+\sqrt[3]{\frac{2}{n... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(3, -6, 9), B(0, -3, 6), C(9, -12, 15)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-3 ;-3-(-6) ; 6-9)=(-3 ; 3 ;-3)$
$\overrightarrow{A C}=(9-3 ;-12-(-6) ; 15-9)=(6 ;-6 ; 6)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{a... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; 5 ;-7) \)
\[
\begin{aligned}
& A_{2}(-3 ; 6 ; 3) \\
& A_{3}(-2 ; 7 ; 3) \\
& A_{4}(-4 ; 8 ;-12)
\end{a... | ## 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 ge... | 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:
$$
\begin{aligned}
& a \times b=(p-3 q) \times(p+2 q)=p \times p+2 \cdot p \times q-3 \cd... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ; 5 ; 2\}$
$b=\{-1 ; 1 ;-1\}$
$c=\{1 ; 1 ; 1\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
(a, b, c)=\left|\begin{array}{ccc}
1 & 5 & 2 \\
-1 & 1 & -1 \\
1 & 1 & 1
\end{array}\right|=
$$
$=1 \cdot\left|\begin{array}{cc}1 &... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$$
\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 &... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-1 ; 1 ; 1)$
$a: 3 x-y+2 z+4=0$
$k=\frac{1}{2}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+2 z+2=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\frac{\ln \left(1+2 x^{2}+x^{3}\right)}{x}, x \neq 0 ; \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}$ | Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+2}-\sqrt[3]{n^{3}+2}\right)}{\frac{1}{n}\left(\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sq... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)\left(\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}\right)^{2}+\sqrt[3]{n^{2... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+3 x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+3 x... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sin ^{2} x-\tan^{2} x}{x^{4}}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \sin x \sim x, \text { as } x \rightarrow 0 \\
& \operatorname{tg} x \sim x, \text { as } x \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin ^{2} x-\operatorname{tg}^{2} x}{x^{4}... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}=\left(e^{0}+0\right)^{\cos 0^{4}}= \\
& =(1+0)^{\cos 0}=1^{1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 18-25 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2}(\cos \pi x)^{\tan(x-2)}$ | ## Solution
$\lim _{x \rightarrow 2}(\cos \pi x)^{\operatorname{tg}(x-2)}=(\cos (\pi \cdot 2))^{\operatorname{tg}(2-2)}=(\cos 2 \pi)^{\operatorname{tg} 0}=1^{0}=1$
## Problem Kuznetsov Limits 20-25 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\ln \left(1-\sin \left(x^{3} \sin \frac{1}{x}\right)\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\arcsin (\sin t) \\
y=\arccos (\cos t)
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\arcsin (\sin t))^{\prime}=t^{\prime}=1$
$y_{t}^{\prime}=(\arccos (\cos t))^{\prime}=t^{\prime}=1$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1}{1}=1$
## Kuznetsov Differentiation 16-4 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in the Cartesian coordinate system.
$$
y=2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}, \frac{1}{4} \leq x \leq 1
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}\rig... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=-\arccos \sqrt{x}+\sqrt{x-x^{2}}, 0 \leq x \leq \frac{1}{4}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(-\arccos \sqrt{x}+\sqrt{x-x^{2}}\righ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
x+\arcsin \left(x^{2} \sin \frac{6}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{\cos x-\cos 3 x}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=\cos t+\sin t \\ y=\sin 2 t\end{array}\right.$ | ## Solution
$x_{t}^{\prime}=(\cos t+\sin t)^{\prime}=-\sin t+\cos t$
$y_{t}^{\prime}=(\sin 2 t)^{\prime}=2 \cos 2 t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 \cos 2 t}{-\sin t+\cos t}=2 \cdot \frac{\cos ^{2} t-\sin ^{2} t}{\cos t-\sin t}=2(\sin t+\cos t) \\
& \lef... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}-3 n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\rig... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}=\lim _{n \rightarrow \infty} \frac{(n-1)!(n+n(n+1)(n+2))}{(n-1)!(1+n(n+1)(n+2))}= \\
& =\lim _{n \rightarrow \infty} \frac{n+n(n+1)(n+2)}{1+n(n+1)(n+2)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}(n+n(n+1)(n+2))}{\frac... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{1^{3}+3 \cdot 1^{2} \cdot x+3 \cdot 1 \cdot x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=$
$=\lim _{x \rightarrow 0} \frac{1+3 x+3 x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=\lim _{x ... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}=\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\ln \left(1+x^{2}\right) \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{x^{2}}{1-\sqr... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{\sin 2 x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$
$=\frac{\lim _{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}=\left(6-\frac{5}{\cos 0}\right)^{\operatorname{tg}^{2} 0}= \\
& =\left(6-\frac{5}{1}\right)^{0^{2}}=1^{0}=1
\end{aligned}
$$
## Problem Kuznetsov Limits $18-28$ | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2} \\
& x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0
\end{aligned}
$$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}=\lim _{y \rightarrow 0}\le... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}
$$ | ## Solution
$\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}=(\sqrt[3]{1}+1-1)^{\sin \left(\frac{\pi \cdot 1}{4}\right)}=(1)^{\frac{\sqrt{2}}{2}}=1$
## Problem Kuznetsov Limits 20-28 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sin t ; d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=\frac{1}{\sqrt{2}} \Rightarrow t=\arcsin \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ;-1 ; 4\}$
$b=\{1 ; 0 ; 3\}$
$c=\{1 ;-3 ; 8\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}1 & -1 & 4 \\ 1 & 0 & 3 \\ 1 & -3 & 8\end{array}\right|=$
$=1 \cdot\left|\begin{array}{cc}0 & 3 \\ -... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}\right)}{\frac{1}{n}\left(\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}\right)}= \\
& =\lim _{n \rightarrow \infty} \f... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3) \cdot(2 n+2)!-(2 n+2)!}=$
$$
=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!((2 n+3)-1)}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!\cd... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}$ | ## Solution
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}+2 x+1\right)}{x-2}=$
$=\lim _{x \rightarrow 2}\left(x^{2}+2 x+1\right)=2^{2}+2 \cdot 2+1=4+4+1=9$
## Problem Kuznetsov Limits 10-20 | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)}{2 \sin x-\tan x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\frac{1}{x}(2 \sin x-\tan x)}=$
$=\frac{... | 7 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}$ | ## Solution
Since $\sin \left(\frac{1}{x}\right)_{\text {- is bounded as }} x \rightarrow 0$, then
$x \sin \left(\frac{1}{x}\right) \rightarrow 0 \quad$ as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}=\frac{2+\ln (e+0)}{\cos 0+\sin 0}=\f... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{2} ; \frac{1}{3} ; 1\right)$
$a: 2 x-3 y+3 z-2=0$
$k=1.5$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 2 x-3 y+3 z-3=0$
Substitute the coordinates of point $A$ into the equ... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+6}-\sqrt{n^{2}-5}\right)}{\frac{1}{n}\left(\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \fra... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{7^{n}}\left(2^{n}+7^{n}\right)}{\frac{1}{7^{n}}\left(2^{n}-7^{n-1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\frac{2}{7}\right)^{n}+1}{\left(\frac{2}{7}\right)^{n}-\... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}=\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt[4]{x}-2)^{2}(\sqrt[4]{x}+2)^{2}}}= \\
& =\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{(\sqrt[4]{x}-2)^{\frac{2}{3}} \sqrt[3]{(\sqrt[4]{x}+2)^{2}}}=\l... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}=\left(\lim _{x \rightarrow 0} \tan\left(\frac{\pi}{4}-x\right)\right)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}= \\
& =\left(\tan\left(\frac{\pi}{4}-0\right)\right)^{\lim _{x \righta... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}=\left(\cos \frac{\pi}{2}+1\right)^{\sin \frac{\pi}{2}}=(0+1)^{1}=1^{1}=1$
## Problem Kuznetsov Limits 20-27 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{9 x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0):$
$f(x)=\left\{\begin{array}{c}\tan\left(x^{3}+x^{2} \sin \left(\frac{2}{x}\right)\right), x \neq 0 \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\frac{4 x-x^{2}}{4}, x_{0}=2$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(\frac{4 x-x^{2}}{4}\right)^{\prime}=\frac{4-2 x}{4}=\frac{2-x}{2}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2-x_{0}}{2}=\frac{2-2}{2}=0$
Since $y^{\prime}\left(x_{0}\right)=0$, the equation of the normal line is:
$x=x_{0}$
$x=2$
Thus, th... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
3^{x^{2} \sin \frac{2}{x}}-1+2 x, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{3}+3 n^{2}+3 n+1-n^{3}+3 n^{2}-3 n+1}{n^{2}+2 n+1+n^{2}-2 n+1}=$
$=\lim _{n \rightarrow \infty} \frac{6 n^{2}+2}{2 n^{2}+2}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 n^{2}+1\... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}\right)}{\frac{1}{n^{2}}(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}= \\
& =\lim _{n \rightarrow \infty}... | -4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}= \\
& =\{1-2=3-4=\ldots=(2 n-1)-2 n=-1\}= \\
& =\lim _{n \rightarrow \infty} \frac{-1 \cdot n}{\sqrt[3]{n^{3}+2 n+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-1) \cdot n}{\frac{1}{n} \sqr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)}{x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}\left(x+\sin x^{2}... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\lim _{x \rightarrow 0} \frac{1}{x+6}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{0+6}}=\left(\lim _{x \rightarrow 0} \frac... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2; -1; 1)$
$a: x-2y+6z-10=0$
$k=\frac{3}{5}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-2 y+6 z-6=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}\right)}{\frac{1}{n^{2}}(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}= \\
& =\lim _{n \rightarrow \... | -3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot \frac{(1+(2 n-1)) n}{2}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot n^{2}=\lim _{n \rightarrow \infty} \frac{1}{\frac{(1+n)... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}$ | Solution
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)^{2}}{\left(x^{3}-x^{2}+x+1\right)(x+1)}=$
$=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)}{x^{3}-x^{2}+x+1}=\frac{\left((-1)^{2}-(-... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)}{\operatorname{arctg} x-x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}\left... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}=\lim _{x \rightarrow 0} \frac{\left(e^{\alpha x}-1\right)-\left(e^{\beta x}-1\right)}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
$=\lim _{x \rightarrow 0} \frac{e^{\alpha x}-1}{2 \sin \frac{x(\alpha... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}=\left(\frac{0^{2}+4}{0+2}\right)^{0^{2}+3}=$
$=\left(\frac{4}{2}\right)^{3}=2^{3}=8$
## Problem Kuznetsov Limits 18-6 | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
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