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## Problem Statement
Find the derivative.
$y=\frac{4 x+1}{16 x^{2}+8 x+3}+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{4 x+1}{\sqrt{2}}$ | ## Solution
$y^{\prime}=\left(\frac{4 x+1}{16 x^{2}+8 x+3}+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{4 x+1}{\sqrt{2}}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{4 \cdot\left(16 x^{2}+8 x+3\right)-(4 x+1) \cdot(32 x+8)}{\left(16 x^{2}+8 x+3\right)^{2}}+\frac{1}{\sqrt{2}} \cdot \frac{1}{1+\left(\frac{4 x+1... | \frac{16}{(16x^{2}+8x+3)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,419 |
Condition of the problem
Find the derivative.
$$
y=4 \ln \frac{x}{1+\sqrt{1-4 x^{2}}}-\frac{\sqrt{1-4 x^{2}}}{x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(4 \ln \frac{x}{1+\sqrt{1-4 x^{2}}}-\frac{\sqrt{1-4 x^{2}}}{x^{2}}\right)^{\prime}= \\
& =4 \cdot \frac{1+\sqrt{1-4 x^{2}}}{x} \cdot\left(\frac{x}{1+\sqrt{1-4 x^{2}}}\right)^{\prime}-\frac{\frac{1}{2 \sqrt{1-4 x^{2}}} \cdot(-8 x) \cdot x^{2}-\sqrt{1-4 x^{2}} \cdot 2 x}{... | \frac{2}{x^{3}\sqrt{1-4x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,420 |
## Problem Statement
Find the derivative.
$y=x \cdot \cos \alpha+\sin \alpha \cdot \ln \sin (x-\alpha)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=(x \cdot \cos \alpha+\sin \alpha \cdot \ln \sin (x-\alpha))^{\prime}=\cos \alpha+\sin \alpha \cdot \frac{1}{\sin (x-\alpha)} \cdot \cos (x-\alpha)= \\
& =\frac{1}{\sin (x-\alpha)} \cdot(\cos \alpha \cdot \sin (x-\alpha)+\sin \alpha \cdot \cos (x-\alpha))=\frac{1}{\sin (x-\al... | \frac{\sinx}{\sin(x-\alpha)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,421 |
The problem statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\sqrt{1-t^{2}} \\
y=\tan \sqrt{1+t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\sqrt{1-t^{2}}\right)^{\prime}=\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)=-\frac{t}{\sqrt{1-t^{2}}}$
$y_{t}^{\prime}=(\tan \sqrt{1+t})^{\prime}=\frac{1}{\cos ^{2}(\sqrt{1+t})} \cdot \frac{1}{2 \sqrt{1+t}}=\frac{1}{2 \cos ^{2}(\sqrt{1+t}) \sqrt{1+t}}$
We obtain:
$$
\begin{aligned}
& y_{... | -\frac{\sqrt{1-}}{2\cdot\cos^{2}(\sqrt{1+})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,422 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\sqrt{3} \cdot \cos t \\
y=\sin t
\end{array}\right. \\
& t_{0}=\frac{\pi}{3}
\end{aligned}
$$ | ## Solution
Since $t_{0}=\frac{\pi}{3}$, then
$x_{0}=\sqrt{3} \cdot \cos \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
$y_{0}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
Let's find the derivatives:
$x_{t}^{\prime}=(\sqrt{3} \cdot \cos t)^{\prime}=-\sqrt{3} \cdot \sin t$
$y_{t}^{\prime}=(\sin t)^{\prime}=\cos t$
$y_{x}^{\prime}=\... | -\frac{x}{3}+\frac{2\sqrt{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,423 |
## Task Condition
Find the $n$-th order derivative.
$y=\sin 2x + \cos (x+1)$ | ## Solution
$y^{\prime}=(\sin 2 x+\cos (x+1))^{\prime}=2 \cos 2 x-\sin (x+1)$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(2 \cos 2 x-\sin (x+1))^{\prime}=-4 \sin 2 x-\cos (x+1)$
$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=(-4 \sin 2 x-\cos (x+1))^{\prime}=-8 \cos 2 x+\sin (x+1)$
$y^{(... | y^{(n)}=2^{n}\cdot\sin(\frac{\pi}{2}\cdotn+2x)+\cos(\frac{\pi}{2}\cdotn+x+1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,424 |
## Task Condition
Find the derivative of the specified order.
$y=\left(3-x^{2}\right) \ln ^{2} x, y^{\prime \prime \prime}=?$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\left(3-x^{2}\right) \ln ^{2} x\right)^{\prime}=-2 x \cdot \ln ^{2} x+\left(3-x^{2}\right) \cdot 2 \cdot \ln x \cdot \frac{1}{x}= \\
& =-2 x \cdot \ln ^{2} x+\frac{2 \cdot\left(3-x^{2}\right) \cdot \ln x}{x}
\end{aligned}
$$
$$
\begin{aligned}
& y^{\prime \prime}=\lef... | \frac{-4\cdot\lnx}{x}+\frac{6\lnx-7x^{2}-15}{x^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,425 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sqrt{1-t^{2}} \\
y=\frac{1}{t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\sqrt{1-t^{2}}\right)^{\prime}=\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)=-\frac{t}{\sqrt{1-t^{2}}}$
$y_{t}^{\prime}=\left(\frac{1}{t}\right)^{\prime}=-\frac{1}{t^{2}}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{1}{t^{2}}\right) /\... | \frac{3-2^{2}}{^{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,426 |
## Problem Statement
Show that the function $y$ satisfies equation (1).
$y=\frac{\sin x}{x}$
$x y^{\prime}+x y=\cos x$ | ## Solution
$y^{\prime}=\left(\frac{\sin x}{x}\right)^{\prime}=\frac{\cos x \cdot x-\sin x}{x^{2}}$
Substitute into equation (1):
$x \cdot \frac{\cos x \cdot x-\sin x}{x^{2}}+\frac{\sin x}{x}=\cos x$
Simplify:
$\cos x \cdot x-\sin x+\sin x=\cos x \cdot x$
$0=0$
The equality holds. The function $y_{\text{satisfie... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,427 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{5 ;-1 ;-2\}$
$b=\{6 ; 0 ; 7\}$
$c_{1}=3 a-2 b$
$c_{2}=4 b-6 a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{2}=-2(3 a-2 b)=-2 c_{1}$ for any $a$ and $b$.
That is, $c_{1}=-\frac{1}{2} \cdot c_{2}$, which means the ve... | c_{1}=-\frac{1}{2}\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,429 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(2 ; 2 ; 7), B(0 ; 0 ; 6), C(-2 ; 5 ; 7)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-2 ; 0-2 ; 6-7)=(-2 ;-2 ;-1)$
$\overrightarrow{A C}=(-2-2 ; 5-2 ; 7-7)=(-4 ; 3 ; 0)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overright... | \frac{2}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,430 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=7 p+q$
$b=p-3 q$
$|p|=3$
$|q|=1$
$(\widehat{p, q})=\frac{3 \pi}{4}$ | ## 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=(7 p+q) \times(p-3 q)=7 \cdot p \times p+7 \cdot(-3) \cdot p \times q+q \times p-... | 33\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,431 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{4 ;-1 ;-6\}$
$b=\{1 ;-3 ;-7\}$
$c=\{2 ;-1 ;-4\}$ | ## 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.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
4 & -1 & -6 \\
1 & -3 & -7 \\
2 & -1 & -4
\end{array}\right|= \\
& =4 \cdot\lef... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,432 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2; -1; -2)$
$M_{2}(1; 2; 1)$
$M_{3}(5; 0; -6)$
$M_{0}(14; -3; 7)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-2 & y-(-1) & z-(-2) \\
1-2 & 2-(-1) & 1-(-2) \\
5-2 & 0-(-1) & -6-(-2)
\end{array}\right|=0
$$
Perform transformations:
$$
\left|\begin{array}{ccc}
x-2 & y+1 & z+2 \\
-1 & 3 & 3 \\
3 & 1 & -4... | 4\sqrt{14} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,434 |
## problem statement
Write the equation of the plane passing through point $A$ and perpendicular to the vector $\overrightarrow{B C}$.
$A(0; -8; 10)$
$B(-5; 5; 7)$
$C(-8; 0; 4)$ | ## Solution
Let's find the vector $\overrightarrow{B C}$:
$\overrightarrow{B C}=\{-8-(-5) ; 0-5 ; 4-7\}=\{-3 ;-5 ;-3\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& -3 \cdot(x-0... | 3x+5y+3z+10=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,435 |
## Task Condition
Find the angle between the planes:
$5 x+3 y+z-18=0$
$2 y+z-9=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{5 ; 3 ; 1\}$
$\overrightarrow{n_{2}}=\{0 ; 2 ; 1\}$
The angle $\phi_{\text {between the planes is determined by the formula: }}$
$$
\begin{aligned}
& \... | \arccos\frac{\sqrt{7}}{5}\approx58^{0}3^{\}7^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,436 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(4 ; 0 ; 5)$
$C(5 ; 4 ; 2)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(4-x)^{2}+(0-0)^{2}+(5-0)^{2}}=\sqrt{16-8 x+x^{2}+0+25}=\sqrt{x^{2}-8 x+41} \\
& A C=\sqrt{(5-x)^{2}+(4-0)^{2}+(2-0)^{2}}=\sqrt{25-10 x+x^{2}+16+4}=\sqrt{x^{2}-10 x+45}
\end{aligned}
$$
Since according to the problem $A B=A C$, then... | A(2;0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,437 |
## 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 $a$?
$A(4 ; 0 ;-3)$
$a: 7 x-y+3 z-1=0$
$k=3$ | ## 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 $a$:
$a^{\prime}: 7 x-y+3 z-3=0$
Substitute the coordinates of point $A$ into the equat... | 16\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,438 |
## Problem Statement
Write the canonical equations of the line.
\[
\begin{aligned}
& x-y+z-2=0 \\
& x-2 y-z+4=0
\end{aligned}
\] | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}_{\text {- its direction }}$ vector.
Since the line belongs to both planes simultaneously, its dir... | \frac{x-8}{3}=\frac{y-6}{2}=\frac{z}{-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,439 |
Condition of the problem
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x-1}{7}=\frac{y-2}{1}=\frac{z-6}{-1} \\
& 4 x+y-6 z-5=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{7}=\frac{y-2}{1}=\frac{z-6}{-1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+7 t \\
y=2+t \\
z=6-t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$4(1+7 t)+(2+t)-6(6-t)-5=0$
$4+28 t+2+t-3... | (8;3;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,440 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(1 ; 2 ; 3)$
$2 x+10 y+10 z-1=0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{2 ; 10 ; 10\}$
Then the equation of the desired line is:
$... | M^{\}(0;-3;-2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,441 |
## 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 | 46,442 |
## Task Condition
Derive the equation of the normal to the given curve at the point with abscissa \( x_{0} \).
\[
y=2 x^{2}-3 x+1, x_{\bar{u}}=1
\] | ## Solution
Let's find $y^{\prime}: y^{\prime}=\left(2 x^{2}-3 x+1\right)^{\prime}=4 x-3$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=4 x_{\overline{0}}^{\prime}-3=4 \cdot 1-3=1$
Since $z^{\prime}\left(x_{0}\right) \neq 0$, the equation of the normal line is:
$y-y_{\overline{\bar{\omega}}}=-\frac{1}{y_{\ove... | -x+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,443 |
## Task Condition
Find the differential $d y$.
$$
y=\ln \left(\cos ^{2} x+\sqrt{1+\cos ^{4} x}\right)
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left(\cos ^{2} x+\sqrt{1+\cos ^{4} x}\right)\right)^{\prime} d x= \\
& =\frac{1}{\cos ^{2} x+\sqrt{1+\cos ^{4} x}} \cdot\left(2 \cdot \cos x \cdot(-\sin x)+\frac{1}{2 \sqrt{1+\cos ^{4} x}} \cdot 4 \cos ^{3} x \cdot(-\sin x)\right) d x= \\
& =\frac{-\... | -\frac{\sin2x\cdot}{\sqrt{1+\cos^{4}x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,444 |
## Task Condition
Find the derivative.
$$
y=\frac{4+3 x^{3}}{x \cdot \sqrt[3]{\left(2+x^{5}\right)^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{3 x^{-3}}{x \cdot \sqrt[3]{\left(2+x^{3}\right)^{2}}}\right)^{\prime}=\left(\frac{2\left(2+x^{3}\right)+x^{-3}}{x \cdot \sqrt[3]{\left(2+x^{-3}\right)^{2}}}\right)^{\prime}= \\
& =\left(\frac{2 \sqrt[3]{2+x^{3}}}{x}+\frac{x^{2}}{\sqrt[3]{\left(2+x^{3}\right)^{2}}... | \frac{-8}{x^{2}(2+x^{3})\sqrt[3]{(2+x^{3})^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,446 |
## Task Condition
Find the derivative.
$$
y=\frac{2\left(\sqrt{2^{x}-1}-\operatorname{arctg} \sqrt{2^{x}-1}\right)}{\ln 2}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2\left(\sqrt{2^{x}-1}-\operatorname{arctg} \sqrt{2^{x}-1}\right)}{\ln 2}\right)^{\prime}= \\
& =\frac{2}{\ln 2} \cdot\left(\frac{1}{2 \sqrt{2^{x}-1}} \cdot 2^{x} \cdot \ln 2-\frac{1}{1+\left(2^{x}-1\right)} \cdot \frac{1}{2 \sqrt{2^{x}-1}} \cdot 2^{x} \cdot \ln 2... | \sqrt{2^{x}-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,447 |
## Task Condition
Find the derivative.
$y=\ln \frac{x^{2}}{1-x^{2}}$ | ## Solution
$y^{\prime}=\left(\ln \frac{x^{2}}{1-x^{2}}\right)^{\prime}=\frac{1-x^{2}}{x^{2}} \cdot\left(\frac{x^{2}}{1-x^{2}}\right)^{\prime}=$
$=\frac{1-x^{2}}{x^{2}} \cdot \frac{2 x \cdot\left(1-x^{2}\right)-x^{2} \cdot(-2 x)}{\left(1-x^{2}\right)^{2}}=$
$=\frac{2 x-2 x^{3}+2 x^{3}}{x^{2} \cdot\left(1-x^{2}\right... | \frac{2}{x\cdot(1-x^{2})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,448 |
## Task Condition
Find the derivative.
$y=\operatorname{ctg}(\cos 2)+\frac{1}{6} \cdot \frac{\sin ^{2} 6 x}{\cos 12 x}$ | ## Solution
$y^{\prime}=\left(\operatorname{ctg}(\cos 2)+\frac{1}{6} \cdot \frac{\sin ^{2} 6 x}{\cos 12 x}\right)^{\prime}=0+\frac{1}{6} \cdot\left(\frac{\sin ^{-2} 6 x}{\cos 12 x}\right)^{\prime}=$
$=\frac{1}{6} \cdot \frac{\left(\sin ^{2} 6 x\right)^{\prime} \cdot \cos 12 x-\sin ^{2} 6 x \cdot(\cos 12 x)^{\prime}}{... | \frac{\sin12x}{\cos^{2}12x}=\frac{\operatorname{tg}12x}{\cos12x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,449 |
## Problem Statement
Find the derivative.
$$
y=\frac{(1+x) \operatorname{arctg} \sqrt{x}}{x^{2}}+\frac{1}{3 x \sqrt{x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{(1+x) \operatorname{arctg} \sqrt{x}}{x^{2}}+\frac{1}{3 x \sqrt{x}}\right)^{\prime}=\left(\frac{(1+x) \operatorname{arctg} \sqrt{x}}{x^{2}}+\frac{1}{3} \cdot x^{-\frac{3}{2}}\right)^{\prime}= \\
& =\frac{((1+x) \operatorname{arctg} \sqrt{x})^{\prime} \cdot x^{2}-(... | -\frac{(2+x)\cdot\operatorname{arctg}\sqrt{x}}{x^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,450 |
## Problem Statement
Find the derivative.
$$
y=\operatorname{arctan} \frac{\sqrt{\operatorname{sinh} 2 x}}{\operatorname{cosh} x-\operatorname{sinh} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{\sqrt{\operatorname{sh} 2 x}}{\operatorname{ch} x-\operatorname{sh} x}\right)^{\prime}=\frac{1}{1+\left(\frac{\sqrt{\operatorname{sh} 2 x}}{\operatorname{ch} x-\operatorname{sh} x}\right)^{2}} \cdot\left(\frac{\sqrt{\operatorname{sh} 2 x}}{\o... | \frac{\operatorname{ch}x+\operatorname{sh}x}{\sqrt{\operatorname{sh}2x}\cdot\operatorname{ch}2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,451 |
## Task Condition
Find the derivative.
$$
y=(\tan x)^{4 e^{r}}
$$ | ## Solution
$$
\begin{aligned}
& y=(\tan x)^{4 e^{-x}} \\
& \ln y=4 e^{x} \cdot \ln (\tan x) \\
& \frac{y^{\prime}}{y}=\left(4 e^{x} \cdot \ln (\tan x)\right)^{\prime}=\left(4 e^{-x}\right)^{\prime} \cdot \ln (\tan x)+4 e^{-x} \cdot(\ln (\tan x))^{\prime}= \\
& =4 e^{x} \cdot \ln (\tan x)+4 e^{x} \cdot \frac{1}{\tan x... | (\tanx)^{4e^{-x}}\cdot4e^{x}\cdot(\ln(\tanx)+\frac{2}{\sin2x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,452 |
## Problem Statement
Find the derivative.
$$
y=\ln \left(4 x-1+\sqrt{16 x^{2}-8 x+2}\right)-\sqrt{16 x^{2}-8 x+2} \cdot \operatorname{arctg}(4 x-1)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \left(4 x-1+\sqrt{16 x^{2}-8 x+2}\right)-\sqrt{16 x^{2}-8 x+2} \cdot \operatorname{arctg}(4 x-1)\right)^{\prime}= \\
& =\frac{1}{4 x-1+\sqrt{16 x^{2}-8 x+2}} \cdot\left(4+\frac{1}{2 \sqrt{16 x^{2}-8 x+2}} \cdot(32 x-8)\right)- \\
& -\left(\frac{1}{2 \sqrt{16 x^{2}-... | \frac{4(1-4x)}{\sqrt{16x^{2}-8x+2}}\cdot\operatorname{arctg}(4x-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,453 |
## Task Condition
Find the derivative.
$$
y=\ln \left(x+\sqrt{x^{2}+1}\right)-\frac{\sqrt{1+x^{2}}}{x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \left(x+\sqrt{x^{2}+1}\right)-\frac{\sqrt{1+x^{2}}}{x}\right)^{\prime}= \\
& =\frac{1}{x+\sqrt{x^{2}+1}} \cdot\left(1+\frac{1}{2 \sqrt{x^{2}+1}} \cdot 2 x\right)-\frac{\frac{1}{2 \sqrt{1+x^{2}}} \cdot 2 x \cdot x-\sqrt{1+x^{2}} \cdot 1}{x^{2}}=
\end{aligned}
$$
$$... | \frac{\sqrt{x^{2}+1}}{x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,454 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{a\left(1+a^{2}\right)}\left(\operatorname{arctg}(a \cos x)+a \ln \left(\operatorname{tg} \frac{x}{2}\right)\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{a\left(1+a^{2}\right)}\left(\operatorname{arctg}(a \cos x)+a \ln \left(\operatorname{tg} \frac{x}{2}\right)\right)\right)^{\prime}= \\
& =\frac{1}{a\left(1+a^{2}\right)}\left(\frac{1}{1+(a \cos x)^{2}} \cdot(a \cos x)^{\prime}+a \frac{1}{\operatorname{tg} \fra... | \frac{\cosx\cdot\operatorname{ctg}x}{1+^{2}\cdot\cos^{2}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,455 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\operatorname{arctg} e^{\frac{t}{2}} \\
y=\sqrt{e^{t}+1}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\operatorname{arctg} e^{\frac{t}{2}}\right)^{\prime}=\frac{1}{1+e^{t}} \cdot e^{\frac{t}{2}} \cdot \frac{1}{2}=\frac{e^{t / 2}}{2\left(1+e^{t}\right)} \\
& y_{t}^{\prime}=\left(\sqrt{e^{t}+1}\right)^{\prime}=\frac{1}{2 \sqrt{e^{t}+1}} \cdot e^{t}=\frac{e^{t}}{2 \sq... | \sqrt{e^{2}+e^{}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,456 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=2 \ln (\operatorname{ctg} t) + \operatorname{ctg} t \\
y=\operatorname{tg} t + \operatorname{ctg} t
\end{array}
\right.
\]
$t_{0}=\frac{\pi... | ## Solution
Since $t_{0}=\frac{\pi}{4}$, then
$x_{0}=2 \ln \left(\operatorname{ctg} \frac{\pi}{4}\right)+\operatorname{ctg} \frac{\pi}{4}=2 \ln 1+1=1$
$y_{0}=\operatorname{tg} \frac{\pi}{4}+\operatorname{ctg} \frac{\pi}{4}=1+1=2$
Let's find the derivatives:
$x_{t}^{\prime}=(2 \ln (\operatorname{ctg} t)+\operatorna... | 2,\;1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,457 |
## Task Condition
Find the $n$-th order derivative.
$y=\sqrt{x}$ | ## Solution
$y^{\prime}=(\sqrt{x})^{\prime}=\left(x^{\frac{1}{2}}\right)^{\prime}=\frac{1}{2} \cdot x^{-\frac{1}{2}}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{2} \cdot x^{-\frac{1}{2}}\right)^{\prime}=\frac{1}{2} \cdot\left(-\frac{1}{2}\right) \cdot x^{-\frac{3}{2}}=-\frac{1}{2^{2}} \cdot x^... | y^{(n)}=(-1)^{n-1}\cdot\frac{\prod_{k=1}^{n-1}(2k-1)}{2^{(n+1)}}\cdotx^{-\frac{2n-1}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,458 |
## Task Condition
Find the derivative of the specified order.
$y=(2 x+3) \ln ^{2} x, y^{\prime \prime \prime}=?$ | ## Solution
$y^{\prime}=2 \cdot \ln ^{2} x+(2 x+3) \cdot 2 \cdot \ln x \cdot \frac{1}{x}=2 \cdot \ln ^{2} x+\frac{2(2 x+3) \cdot \ln x}{x}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2 \cdot \ln ^{2} x+\frac{2(2 x+3) \cdot \ln x}{x}\right)^{\prime}=4 \cdot \ln x \cdot \frac{1}{x}+\left(4 \ln x+\frac{6 ... | \frac{4\lnx\cdot(3-x)-18}{x^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,459 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\tan t \\
y=\frac{1}{\sin 2 t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\tan t)^{\prime}=\frac{1}{\cos ^{2} t}$
$y_{t}^{\prime}=\left(\frac{1}{\sin 2 t}\right)^{\prime}=-\frac{1}{\sin ^{2} 2 t} \cdot \cos 2 t \cdot 2=-\frac{2 \cos 2 t}{\sin ^{2} 2 t}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{2 \cos 2 ... | -\frac{2\cdot\cos^{3}}{\sin\cdot\cos2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,460 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(3-4 n)^{2}}{(n-3)^{3}-(n+3)^{3}}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(3-4 n)^{2}}{(n-3)^{3}-(n+3)^{3}}=$
$\lim _{n \rightarrow x} \frac{9-24 n+16 n^{2}}{n l^{3}-3 \cdot n l^{2} \cdot 3+3 \cdot n \cdot 3^{2}-3^{3}-n^{3}-3 \cdot n^{2} \cdot 3-3 \cdot n \cdot 3^{2}-3^{3}}=$
$=\lim _{n \rightarrow x} \frac{9-\cdot 24 n+16 n^{2}}{-3 \cdot n^... | -\frac{8}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,462 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\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 | 46,465 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1+4+7+\ldots+(3 n-2)}{\sqrt{5 n^{4}+n+1}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+4+7+\ldots+(3 n-2)}{\sqrt{5 n^{4}+n+1}}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{\sqrt{5 n^{4}+n+1}} \cdot \frac{(1+(3 n-2)) n}{2}=\lim _{n \rightarrow \infty} \frac{n(3 n-1)}{2 \sqrt{5 n^{4}+n+1}}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac... | \frac{3}{2\sqrt{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,466 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow 2} \frac{3 x^{2}-5 x-2}{x-2}=7$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | \delta(\varepsilon)=\frac{\varepsilon}{3} | Calculus | proof | Yes | Yes | olympiads | false | 46,468 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=-5 x^{2}-8, x_{0}=2$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\right|<\del... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,469 |
## 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 | 46,470 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+x^{2}}-(1+x)}{x}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+x^{2}}-(1+x)}{x}=\left\{\frac{0}{0}\right\}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+x^{2... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,471 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\arcsin 3 x}{\sqrt{2+x}-\sqrt{2}}
$$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\arcsin 3 x \sim 3 x$, as $x \rightarrow 0 (3 x \rightarrow 0)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\arcsin 3 x}{\sqrt{2+x}-\sqrt{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{3 x}{\sqrt{2+x}-\sqrt{2}}= \\
... | 6\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,472 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sqrt{x^{2}-x+1}-1}{\tan \pi x}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt{x^{2}-x+1}-1}{\tan \pi x}=\lim _{y \rightarrow 0} \frac{\sqrt{(y+1)^{2}-(y+1)+1}-1}{\tan \pi(y+1)}= \\
& =\lim _{y \rightarrow 0} \frac{\sqrt{y^{2}+2 y+... | \frac{1}{2\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,473 |
## Problem Statement
Calculate the limit of the function:
$$
\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 | 46,474 |
## 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 | 46,475 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{x^{2}\left(e^{x}-e^{-x}\right)}{e^{x^{3}+1}-e}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{x^{2}\left(e^{x}-e^{-x}\right)}{e^{x^{3}+1}-e}=\lim _{x \rightarrow 0} \frac{x^{2} e^{-x}\left(e^{2 x}-1\right)}{e\left(e^{x^{3}}-1\right)}=$
$$
\begin{aligned}
& e^{2 x}-1 \sim 2 x, \text { as } x \rightarrow 0(2 x \rightarrow 0) \\
& e^{x^{3}}-1 \sim x^{3}, \text { as } x ... | \frac{2}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,476 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(2-e^{\arcsin ^{2} \sqrt{x}}\right)^{\frac{3}{x}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(2-e^{\arcsin ^{2} \sqrt{x}}\right)^{\frac{3}{x}}=\lim _{x \rightarrow 0}\left(e^{\ln \left(2-e^{\arcsin ^{2} \sqrt{x}}\right)}\right)^{\frac{3}{x}}= \\
& =\lim _{x \rightarrow 0} e^{\frac{3}{x} \cdot \ln \left(2-e^{\arcsin ^{2} \sqrt{x}}\right)}=\exp \left\... | e^{-3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,477 |
## 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 | 46,478 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow a}\left(2-\frac{x}{a}\right)^{\tan\left(\frac{\pi x}{2 a}\right)}$ | ## Solution
$\lim _{x \rightarrow a}\left(2-\frac{x}{a}\right)^{\operatorname{tg}\left(\frac{\pi x}{2 a}\right)}=\lim _{x \rightarrow a}\left(e^{\ln \left(2-\frac{x}{a}\right)}\right)^{\operatorname{tg}\left(\frac{\pi x}{2 a}\right)}=$
$=\lim _{x \rightarrow a}\left(e^{\operatorname{tg}\left(\frac{\pi x}{2 a}\right) ... | e^{\frac{2}{\pi}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,479 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-x^{2}}{1-x}}$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-x^{2}}{1-x}}=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{(1-x)(1+x)}{1-x}}=$
$=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{1+x}=\left(\frac{1+1}{2+1}\right)^{1+1}=\left(\frac{2}{3}\right)^{2}=\frac{4}{9}$
## Problem Kuz... | \frac{4}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,480 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{7 n-1}{n+1}, a=7$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{7 n-1-7 n-7}{n+1}\right| \\
& \left|\frac{-8}{n+1}\right| \\
& \frac{8}{n+1} \\
& n+1>\frac{8}{\varepsilon} ;=> \\
& n>\left(\frac{8}{\varep... | N(\varepsilon)=[\frac{8}{\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 46,482 |
## Task Condition
Calculate the limit of the numerical sequence:
$$
(6-n)^{2}-(6+n)^{2}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(6-n)^{2}-(6+n)^{2}}{(6+n)^{2}-(1-n)^{2}}=\lim _{n \rightarrow \infty} \frac{36-12 n+n^{2}-36-12 n-n^{2}}{36+12 n+n^{2}-1+2 n-n^{2}}= \\
& =\lim _{n \rightarrow \infty} \frac{-24 n}{35+14 n}=\lim _{n \rightarrow \infty} \frac{-24}{\frac{35}{n}+14}=\fr... | -\frac{12}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,483 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\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 | 46,484 |
## 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}}{\sqrt[3]{n}-n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{3 n-1}-\sqrt[3]{125 n^{3}+n}\right)}{\frac{1}{n}(\sqrt[3]{n}-n)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{3}{n}-\frac{1}{n^{2}}}-... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,485 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{5}-8}-n \sqrt{n\left(n^{2}+5\right)}}{\sqrt{n}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n^{5}-8}-n \sqrt{n\left(n^{2}+5\right)}}{\sqrt{n}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{n^{5}-8}-n \sqrt{n\left(n^{2}+5\right)}\right)\left(\sqrt{n^{5}-8}+n \sqrt{n\left(n^{2}+5\right)}\right)}{\sqrt{n}\left(\sqrt{n^{5}-8}+n \sq... | -\frac{5}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,486 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{\sqrt{9 n^{4}+1}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{\sqrt{9 n^{4}+1}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{9 n^{4}+1}} \cdot \frac{(n+1) n}{2}= \\
& =\lim _{n \rightarrow \infty} \frac{(n+1) n}{2 \sqrt{9 n^{4}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}(n+1) n}{\frac{1... | \frac{1}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,487 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2}{2 n^{2}+1}\right)^{n^{2}}$ | ## Solution
$\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2}{2 n^{2}+1}\right)^{n^{2}}=\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+1+1}{2 n^{2}+1}\right)^{n^{2}}=$
$=\lim _{n \rightarrow \infty}\left(1+\frac{1}{2 n^{2}+1}\right)^{n^{2}}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{2 n^{2}+1}\right)^{\left(2 n^... | \sqrt{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,488 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow-\frac{1}{2}} \frac{6 x^{2}+x-1}{x+\frac{1}{2}}=-5$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=-2 x^{2}-5, x_{0}=2$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\right|<\del... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,490 |
## 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 | 46,491 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{x^{3}+8}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{x^{3}+8}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\left(x^{3}+8\right)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \... | \frac{1}{144} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,492 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{4 x}{\tan(\pi(2+x))}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\operatorname{tg} \pi x \sim \pi x$, as $x \rightarrow 0 (\pi x \rightarrow 0)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{4 x}{\operatorname{tg}(\pi(2+x))}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{4 x}{\operator... | \frac{4}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,493 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\tan^{2} \pi x}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\tan^{2} \pi x}=\lim _{y \rightarrow 0} \frac{1+\cos \pi(y+1)}{\tan^{2} \pi(y+1)}= \\
& =\lim _{y \rightarrow 0} \frac{1+\cos (\pi y+\pi)}{\tan^... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,494 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\tan 2 x}-e^{-\sin 2 x}}{\sin x-1}$ | ## Solution
Substitution:
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\operatorname{tg} 2 x}-e^{-\sin 2 x}}{\sin x-1}=\lim _{y \rightarrow 0} \frac{e^{\operatorname{tg} 2\left(y+\frac{\pi}{2}\right... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,495 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{3^{2 x}-5^{3 x}}{\operatorname{arctg} x+x^{3}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{3^{2 x}-5^{3 x}}{\operatorname{arctg} x+x^{3}}=\lim _{x \rightarrow 0} \frac{\left(9^{x}-1\right)-\left(125^{x}-1\right)}{\operatorname{arctg} x+x^{3}}=$
$=\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 9}\right)^{x}-1\right)-\left(\left(e^{\ln 125}\right)^{x}-1\right)}{\o... | \ln\frac{9}{125} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,496 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{1+\tan x}-\sqrt{1+\sin x})(\sqrt{1+\tan x}+\sqrt{1+\sin x})}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{1+\tan x-(1+\sin x)}{x^{3}(\sqrt{1... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,497 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)^{\operatorname{ctg}^{3} x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)^{\operatorname{ctg}^{3} x}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(\frac{1+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)}\right)^{\operatorname{ctg}^{3} x}=$
$=\lim _{x \rightarro... | e^{\frac{\beta^{2}-\alpha^{2}}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,498 |
## 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 | 46,499 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 8}\left(\frac{2 x-7}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-2}}$ | ## Solution
$\lim _{x \rightarrow 8}\left(\frac{2 x-7}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-2}}=\lim _{x \rightarrow 8}\left(e^{\ln \left(\frac{2 x-7}{x+1}\right)}\right)^{\frac{1}{\sqrt[3]{x}-2}}=$
$=\lim _{x \rightarrow 8} e^{\frac{1}{\sqrt[3]{x}-2} \cdot \ln \left(\frac{2 x-7}{x+1}\right)}=\exp \left\{\lim _{x \right... | e^{\frac{4}{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,500 |
## 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
$$
\begin{aligned}
& \lim _{x \rightarrow 2}\left(\frac{\sin (3 \pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}=\lim _{x \rightarrow 2}\left(\frac{\sin (2 \pi x+\pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}= \\
& =\lim _{x \rightarrow 2}\left(\frac{\sin 2 \pi x \cdot \cos \pi x+\cos 2 \pi x \cdot \sin \pi x}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,501 |
Condition of the problem
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{e^{1 / n}+\sin \left(\frac{n}{n^{2}+1}\right) \cdot \cos n}{1+\cos \left(\frac{1}{n}\right)}$ | ## Solution
Since $\cos n$ is bounded, and $\sin \left(\frac{n}{n^{2}+1}\right) \rightarrow 0$ as $n \rightarrow \infty$, then
$$
\sin \left(\frac{n}{n^{2}+1}\right) \cdot \cos n \rightarrow 0, \text { as } n \rightarrow \infty
$$
Then:
$$
\lim _{n \rightarrow \infty} \frac{e^{1 / n}+\sin \left(\frac{n}{n^{2}+1}\ri... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,502 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{2 n^{3}}{n^{3}-2}, a=2$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{2 n^{3}-2 n^{3}+4}{n^{3}-2}\right| \\
& \left|\frac{4}{n^{3}-2}\right| \\
& \frac{4}{n^{3}-2} \\
& n^{3}-2>\frac{4}{\varepsilon} ;=> \\
& n^... | N(\varepsilon)=[\sqrt[3]{2+\frac{4}{\varepsilon}}]+1 | Calculus | proof | Yes | Yes | olympiads | false | 46,503 |
## 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=\lim _{n \rightarrow \infty} \frac{4 n^{2}+4 n+1-n^{2}-2 n-1}{n^{2}+n+1}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 n^{2}+2 n\right)}{\frac{1}{n^{2}}\left(n^{2}+n+1\right)}=\lim _{n \rightarrow \inf... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,504 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[3]{n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[3]{n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[6]{n}+\sqrt[3]{n^{10}+1}\right)}{\frac{1}{n^{2}}(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}= \\
& =\lim _{n \rightarrow \infty}... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,505 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[3]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[3]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[6]{n}+\sqrt[3]{32 n^{10}+1}\right)}{\frac{1}{n^{2}}(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}= \\
& =\lim _{n \rightarrow \... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,506 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{2}+5\right)\left(n^{4}+2\right)}-\sqrt{n^{6}-3 n^{3}+5}}{n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{2}+5\right)\left(n^{4}+2\right)}-\sqrt{n^{6}-3 n^{3}+5}}{n}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{\left(n^{2}+5\right)\left(n^{4}+2\right)}-\sqrt{n^{6}-3 n^{3}+5}\right)\left(\sqrt{\left(n^{2}+5\right)\left(n^{4}+2\right... | \frac{5}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,507 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{2+4+\ldots+2 n}{n+3}-n\right)$ | $$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2+4+\ldots+2 n}{n+3}-n\right)=\lim _{n \rightarrow \infty}\left(\frac{\left(\frac{(2+2 n) n}{2}\right)}{n+3}-n\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{(1+n) n}{n+3}-n\right)=\lim _{n \rightarrow \infty}\left(\frac{(1+n) n}{n+3}-\frac{n(n+3)}{n+3... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,508 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{4 n^{2}+4 n-1}{4 n^{2}+2 n+3}\right)^{1-2 n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{4 n^{2}+4 n-1}{4 n^{2}+2 n+3}\right)^{1-2 n}=\lim _{n \rightarrow \infty}\left(\frac{4 n^{2}+2 n+3+2 n-4}{4 n^{2}+2 n+3}\right)^{1-2 n}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{2 n-4}{4 n^{2}+2 n+3}\right)^{1-2 n}=\lim _{n \rightarrow \in... | \frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,509 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow \frac{1}{3}} \frac{15 x^{2}-2 x-1}{x-\frac{1}{3}}=8$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | \delta(\varepsilon)=\frac{\varepsilon}{15} | Calculus | proof | Yes | Yes | olympiads | false | 46,510 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=5 x^{2}+5, x_{0}=8$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,511 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3} \frac{x^{3}-4 x^{2}-3 x+18}{x^{3}-5 x^{2}+3 x+9}$ | ## Solution
$\lim _{x \rightarrow 3} \frac{x^{3}-4 x^{2}-3 x+18}{x^{3}-5 x^{2}+3 x+9}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 3} \frac{(x-3)\left(x^{2}-x-6\right)}{(x-3)\left(x^{2}-2 x-3\right)}=$
$=\lim _{x \rightarrow 3} \frac{x^{2}-x-6}{x^{2}-2 x-3}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 3} \frac{... | \frac{5}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,512 |
## Problem Statement
Calculate the limit of the function:
$$
\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 | 46,513 |
## 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:
$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{2 x \cdot x}{\frac{x^{2}}{2}}=\lim... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,514 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}$ | ## Solution
$\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan ^{2} 2 x}=\lim _{x \rightarrow \pi} \frac{-2 \sin \frac{3 x+x}{2} \sin \frac{3 x-x}{2}}{\tan ^{2} 2 x}=$
$=\lim _{x \rightarrow \pi} \frac{-2 \sin 2 x \sin x}{\tan ^{2} 2 x}=$
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,515 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\sin \left(\frac{x^{2}}{\pi}\right)}{2^{\sqrt{\sin x+1}}-2}$ | ## Solution
Substitution:
$$
\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{\sin \left(\frac{x^{2}}{\pi}\right)}{2^{\sqrt{\sin x+1}}-2}=\lim _{y \rightarrow 0} \frac{\sin \left(\frac{(y+\p... | \frac{2}{\ln2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,516 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2^{3 x}-3^{5 x}}{\sin 7 x-2 x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2^{3 x}-3^{5 x}}{\sin 7 x-2 x}=\lim _{x \rightarrow 0} \frac{\left(8^{x}-1\right)-\left(243^{x}-1\right)}{\sin 7 x-2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 8}\right)^{x}-1\right)-\left(\left(e^{\ln 243}\right)^{x}-1\right)}{\sin 7 x-2 x... | \frac{1}{5}\ln\frac{2^{3}}{3^{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,517 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{e^{x}-e}{\sin \left(x^{2}-1\right)}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow 1} \frac{e^{x}-e}{\sin \left(x^{2}-1\right)}=\lim _{y \rightarrow 0} \frac{e^{y+1}-e}{\sin \left((y+1)^{2}-1\right)}=$
$=\lim _{y \rightarrow 0} \frac{e\left(e^{y}-1\right)}{\sin \left(y^... | \frac{e}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,518 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{1+x^{2} \cdot 2^{x}}{1+x^{2} \cdot 5^{x}}\right)^{\frac{1}{\sin ^{3} x}}
$$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{1+x^{2} \cdot 2^{x}}{1+x^{2} \cdot 5^{x}}\right)^{\frac{1}{\sin ^{3} x}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(\left(1+x^{2} \cdot 2^{x}\right) /\left(1+x^{2} \cdot 5^{x}\right)\right)}\right)^{\frac{1}{\sin ^{3} x}}=$
$=\lim _{x \rightarrow 0} e^{\frac{1}{\sin... | \frac{2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,519 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{x^{3}+4}{x^{3}+9}\right)^{\frac{1}{x+2}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{x^{3}+4}{x^{3}+9}\right)^{\frac{1}{x+2}}=\left(\frac{0^{3}+4}{0^{3}+9}\right)^{\frac{1}{0+2}}=\left(\frac{4}{9}\right)^{\frac{1}{2}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$
## Problem Kuznetsov Limits 18-31 | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,520 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{\ln (3+2 x)}{\ln (2-x)}}$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{\ln (3+2 x)}{\ln (2-x)}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{2 x-1}{x}\right)}\right)^{\frac{\ln (3+2 x)}{\ln (2-x)}}=$
$=\lim _{x \rightarrow 1} e^{\frac{\ln (3+2 x)}{\ln (2-x)} \cdot \ln \left(\frac{2 x-1}{x}\right)}=\exp \left\{\li... | \frac{1}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,521 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{e^{2 x}-e^{2}}{x-1}\right)^{x+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{e^{2 x}-e^{2}}{x-1}\right)^{x+1}=\left(\lim _{x \rightarrow 1} \frac{e^{2 x}-e^{2}}{x-1}\right)^{\lim _{x \rightarrow 1} x+1}= \\
& =\left(\lim _{x \rightarrow 1} \frac{e^{2 x}-e^{2}}{x-1}\right)^{1+1}=\left(\lim _{x \rightarrow 1} \frac{e^{2}\left(e^... | 4e^4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,522 |
## 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 | 46,523 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{8 ; 0 ; 5\}$
$p=\{2 ; 0 ; 1\}$
$q=\{1 ; 1 ; 0\}$
$r=\{4 ; 1 ; 2\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | p-2q+2r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,524 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{7 ; 9 ;-2\}$
$b=\{5 ; 4 ; 3\}$
$c_{1}=4 a-b$
$c_{2}=4 b-a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=4 a-b=\{4 \cdot 7-5 ; 4 \cdot 9-4 ; 4 \cdot(-2)-3\}=\{23 ; 32 ;-11\} \\
& c_{2}=4 b-a=\{4 \cdot 5-7 ; 4 \... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,525 |
## 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 | 46,526 |
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