## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(2-e^{\sin x}\right)^{\operatorname{ctg} \pi x}$
## Solution

$\lim _{x \rightarrow 0}\left(2-e^{\sin x}\right)^{\operatorname{ctg} \pi x}=\lim _{x \rightarrow 0}\left(e^{\ln \left(2-e^{\sin x}\right)}\right)^{\operatorname{ctg} \pi x}=$

$=\lim _{x \rightarrow 0} e^{\operatorname{ctg} \pi x \cdot \ln \left(2-e^{\sin x}\right)}=$

$=\exp \left\{\lim _{x \rightarrow 0} \operatorname{ctg} \pi x \cdot \ln \left(2-e^{\sin x}\right)\right\}=$

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\left(1-e^{\sin x}\right)\right) \sim 1-e^{\sin x}$, as $x \rightarrow 0\left(1-e^{\sin x} \rightarrow 0\right)$

$\operatorname{tg} \pi x \sim \pi x$, as $x \rightarrow 0(\pi x \rightarrow 0)$

We get:
$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\pi x} \cdot\left(1-e^{\sin x}\right)\right\}=\exp \left\{\lim _{x \rightarrow 0}-\frac{1}{\pi x} \cdot\left(e^{\sin x}-1\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$e^{\sin x}-1 \sim \sin x$, as $x \rightarrow 0(\sin x \rightarrow 0)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{1}{\pi x} \cdot \sin x\right\}=\exp \left\{\lim _{x \rightarrow 0}-\frac{\sin x}{\pi x}\right\}=$

Using the substitution of equivalent infinitesimals:

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

We get:

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{x}{\pi x}\right\}=\exp \left\{\lim _{x \rightarrow 0}-\frac{1}{\pi}\right\}=e^{-\frac{1}{\pi}}$

## Problem Kuznetsov Limits 17-15