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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 Limi... | 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:
$$
\lim _{x... | 50 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-2 x}}{2 \arcsin x-\sin x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-2 x}}{2 \arcsin x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)}{2 \arcsin x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}(2 \arcsin x-\sin x)}=... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\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 \r... | 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 (\overrigh... | 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 \... | 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 geome... | 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}}{\sqrt[4]{n^{5}+5}+\sqrt{n}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}\right)}{\frac{1}{n}\left(\sqrt[4]{n^{5}+5}+\sqrt{n}\right)}= \\
& =\lim _{n \rightarrow \infty} \f... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-6 x^{2}+12 x-8}{x^{3}-3 x^{2}+4}$ | ## Solution
$$
\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 \ri... | 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... | 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 $\overrightarrow{A C}$:
$$
\be... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{2 x}{x^{2}+1}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
\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}... | 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 ... | 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 equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2 ; 3 ; 1)$
$M_{2}(4 ; 1 ;-2)$
$M_{3}(6 ; 3 ; 7)$
$M_{0}(-5 ;-4 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-2 & y-3 & z-1 \\ 4-2 & 1-3 & -2-1 \\ 6-2 & 3-3 & 7-1\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-3 & z-1 \\
2 & -2 & -3 \\
4 & 0 & 6
\end{... | 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\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}\... | 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)}... | 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{\... | 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 (\overrightarr... | 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 \... | 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 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}
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} \fra... | 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\left((n+1)^{2}-(n-1)^{2}\right) \cdot\left((n+1)^{2}+(n-1)^{2}\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+2 n+1-n^{2}+2 n-1\right) \c... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}+2}-5 n^{2}}{n-\sqrt{n^{4}-n+1}}$ | ## Solution
$$
\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}}... | 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2+4+6+\ldots+2 n}{1+3+5+\ldots+(2 n-1)}=\lim _{n \rightarrow \infty} \frac{\left(\frac{(2+2 n) n}{2}\right)}{\left(\frac{(1+(2 n-1)) n}{2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{(2+2 n) n}{(1+(2 n-1)) n}=\lim _{n \rightarrow \infty} \frac{2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\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... | 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:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{1-\sin \left(\frac{x}{2}\right)}{\pi-x}=\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{y+\pi}{2}\right)}{\pi-(... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\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-\... | -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
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}= \\
& =\lim _{x \rightarrow \frac{\pi}{6}} \frac{2\left(\sin ^{2} x+\frac{1}{2} \cdot \sin x+\frac{1}{16}\right)-\frac{9}{8}}{2\left(\sin ^{2} x-\frac{3}{2} \sin x+\frac{9}{16}\right)-\frac{1}{8... | -3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\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 \righta... | 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^{\... | 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:
$$
\begin{aligned}
& t=\sqrt{\frac{x+3}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{1 \cdot x-(x+3) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{x+3}} \c... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the definite integral:
$$
\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}{... | 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:
$$
\begin{aligned}
& t=\sqrt{\frac{x}{x-1}}, \text { hence } t^{2}=\frac{x}{x-1}, \text { that is } x=1+\frac{1}{t^{2}-1}, \text { and } \\
& d x... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\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}... | 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+1}-\sqrt[3]{n^{3}+1}}{\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+1}-\sqrt[3]{n^{3}+1}\right)}{\frac{1}{n}\left(\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-3} \frac{x^{3}+7 x^{2}+15 x+9}{x^{3}+8 x^{2}+21 x+18}$ | ## Solution
$$
\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\}=\l... | 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
$$
\begin{aligned}
& \lim _{x \rightarrow-8} \frac{10-x-6 \sqrt{1-x}}{2+\sqrt[3]{x}}=\lim _{x \rightarrow-8} \frac{(10-x-6 \sqrt{1-x})(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})}= \\
& =\lim _{x \rightarrow-8} \frac{(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{100-20 x+x^... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-4 ; 2 ; 6)$
$M_{2}(2 ;-3 ; 0)$
$M_{3}(-10 ; 5 ; 8)$
$M_{0}(-12 ; 1 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\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 \... | 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 equatio... | 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+\... | 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}+1}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}+1\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n^{2}-\sqrt{n^{3}+1}}{\sqrt[3]{n^{6}+2}-n}$ | ## Solution
$$
\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^{... | 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
$\lim _{n \rightarrow \infty} n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)=$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)\left(\sqrt{n^{4}+3}+\sqrt{n^{4}-2}\right)}{\sqrt{n^{4}+3}+\sqrt{n^{4}-2}}= \\
& =\lim _{n \rightarrow \infty} \frac{n\left(n^{4}+... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{x^{2}-16}}$ | ## Solution
$$
\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)... | 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 _{... | 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 |
## 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:
$$
\begin{aligned}
& a \times b=(6 p-q) \times(5 q+p)=6 \cdot 5 \cdot p \times q+6 \cdot p \t... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(1 ; 2 ; 2)$
$a: 3x - z + 5 = 0$
$k = -\frac{1}{5}$ | ## Solution
When transforming similarity with the center at the origin of the coordinate plane
$a: A x+B y+C z+D=0$ and the coefficient $k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-z-1=0$
Substitute the coordinates of point $A$ into the eq... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\operatorname{arctg}\left(x^{3}-x^{\frac{3}{2}} \sin \frac{1}{3 x}\right), x \neq 0 ; \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\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:
$$
\begin{aligned}
& \lim _{x \rightarrow 2 \pi} \frac{(x-2 \pi)^{2}}{\operatorname{tg}(\cos x-1)}=\lim _{y \rightarrow 0} \frac{((y+2 \pi)-2 \pi)^{2}}{\operatorname{tg}(\cos (y+2 \pi)-1)}= \\
& =\... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{4 x}-e^{-2 x}}{2 \operatorname{arctan} x-\sin x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{4 x}-e^{-2 x}}{2 \operatorname{arctg} x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)}{2 \operatorname{arctg} x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\tan 4 x}{x}\right)^{2+x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\tan 4 x}{x}\right)^{2+x}=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{\lim _{x \rightarrow 0} 2+x}=$
$=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{2+0}=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{2}=$
Using the substitution of equ... | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\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}}}-... | 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 |
## 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 \... | 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{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 |
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