## 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

$\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)^{\operatorname{ctg}^{3} x}=$

$=\lim _{x \rightarrow 0}\left(e^{\ln \left(\frac{1+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)}\right)^{\operatorname{ctg}^{3} x}=$

$=\lim _{x \rightarrow 0} e^{\operatorname{ctg}^{3} x \cdot \ln \left(\frac{1+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)}=$

$=\exp \left\{\lim _{x \rightarrow 0} \operatorname{ctg}^{3} x \cdot \ln \left(\frac{1+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\operatorname{tg}^{3} x} \cdot \ln \left(\frac{1+\sin x \cdot \cos \beta x-\sin x \cdot \cos \beta x+\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\operatorname{tg}^{3} x} \cdot \ln \left(1-\frac{\sin x \cdot \cos \beta x-\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\frac{\sin x \cdot \cos \beta x-\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right) \sim-\frac{\sin x \cdot \cos \beta x-\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}$, as

$x \rightarrow 0\left(-\frac{\sin x \cdot \cos \beta x-\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{1}{x^{3}} \cdot \frac{\sin x \cdot \cos \beta x-\sin x \cdot \cos \alpha x}{1+\sin x \cdot \cos \beta x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{\sin x}{x^{3}} \cdot \frac{\cos \beta x-\cos \alpha x}{1+\sin x \cdot \cos \beta x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{\sin x}{x^{3}} \cdot \frac{-2 \sin \frac{\beta x+\alpha x}{2} \sin \frac{\beta x-\alpha x}{2}}{1+\sin x \cdot \cos \beta x}\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin x \sim x$, as $x \rightarrow 0$

$\sin \frac{\beta x+\alpha x}{2} \sim \frac{\beta x+\alpha x}{2}, \quad x \rightarrow 0\left(\frac{\beta x+\alpha x}{2} \rightarrow 0\right)$
$\sin \frac{\beta x-\alpha x}{2} \sim \frac{\beta x-\alpha x}{2}, \quad x \rightarrow 0\left(\frac{\beta x-\alpha x}{2} \rightarrow 0\right)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{x}{x^{3}} \cdot \frac{-2 \cdot \frac{\beta x+\alpha x}{2} \cdot \frac{\beta x-\alpha x}{2}}{1+\sin x \cdot \cos \beta x}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{2 \frac{\beta+\alpha}{2} \cdot \frac{\beta-\alpha}{2}}{1+\sin x \cdot \cos \beta x}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{\beta^{2}-\alpha^{2}}{2(1+\sin x \cdot \cos \beta x)}\right\}=$

$=\exp \left\{\frac{\beta^{2}-\alpha^{2}}{2(1+\sin 0 \cdot \cos (\beta \cdot 0))}\right\}=$

$=\exp \left\{\frac{\beta^{2}-\alpha^{2}}{2}\right\}=e^{\frac{\beta^{2}-\alpha^{2}}{2}}$

## Problem Kuznetsov Limits 17-5