## 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

$\lim _{x \rightarrow a}\left(2-\frac{x}{a}\right)^{\operatorname{tg}\left(\frac{\pi x}{2 a}\right)}=\lim _{x \rightarrow a}\left(e^{\ln \left(2-\frac{x}{a}\right)}\right)^{\operatorname{tg}\left(\frac{\pi x}{2 a}\right)}=$

$=\lim _{x \rightarrow a}\left(e^{\operatorname{tg}\left(\frac{\pi x}{2 a}\right) \cdot \ln \left(2-\frac{x}{a}\right)}\right)=\exp \left\{\lim _{x \rightarrow a} \operatorname{tg}\left(\frac{\pi x}{2 a}\right) \cdot \ln \left(2-\frac{x}{a}\right)\right\}=$

Substitution:

$x=y+a \Rightarrow y=x-a$

$x \rightarrow a \Rightarrow y \rightarrow 0$

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \operatorname{tg}\left(\frac{\pi(y+a)}{2 a}\right) \cdot \ln \left(2-\frac{(y+a)}{a}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0} \operatorname{tg}\left(\frac{\pi y}{2 a}+\frac{\pi}{2}\right) \cdot \ln \left(1-\frac{y}{a}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0}-\operatorname{ctg}\left(\frac{\pi y}{2 a}\right) \cdot \ln \left(1-\frac{y}{a}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{\operatorname{tg}\left(\frac{\pi y}{2 a}\right)} \cdot \ln \left(1-\frac{y}{a}\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$\operatorname{tg}\left(\frac{\pi y}{2 a}\right) \sim \frac{\pi y}{2 a}{ }_{\text {as }} y \rightarrow 0\left(\frac{\pi y}{2 a} \rightarrow 0\right)$

$\ln \left(1-\frac{y}{a}\right) \sim-\frac{y}{a}$, as $\quad y \rightarrow 0\left(-\frac{y}{a} \rightarrow 0\right)$

We get:

$=\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{\left(\frac{\pi y}{2 a}\right)} \cdot\left(-\frac{y}{a}\right)\right\}=\exp \left\{\lim _{y \rightarrow 0} \frac{2}{\pi}\right\}=$

$=\exp \left\{\frac{2}{\pi}\right\}=e^{\frac{2}{\pi}}$

## Problem Kuznetsov Limits 19-8