## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow e}\left(\frac{\ln x-1}{x-e}\right)^{\sin \left(\frac{\pi}{2 e} x\right)}$
## Solution

$\lim _{x \rightarrow e}\left(\frac{\ln x-1}{x-e}\right)^{\sin \left(\frac{\pi}{2 e} x\right)}=\lim _{x \rightarrow e}\left(\frac{\ln x-\ln e}{x-e}\right)^{\sin \left(\frac{\pi}{2 e} x\right)}=$

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

$=\left(\lim _{x \rightarrow e} \frac{\ln \frac{x}{e}}{x-e}\right)^{\sin \left(\frac{\pi}{2}\right)}=\left(\lim _{x \rightarrow e} \frac{\ln \frac{x}{e}}{x-e}\right)^{1}=$

$=\lim _{x \rightarrow e} \frac{\ln \frac{x}{e}}{x-e}=$

Substitution:

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

$x \rightarrow e \Rightarrow y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{\ln \frac{y+e}{e}}{(y+e)-e}=\lim _{y \rightarrow 0} \frac{\ln \left(1+\frac{y}{e}\right)}{y}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\frac{y}{e}\right) \sim \frac{y}{e} \text{ as } y \rightarrow 0\left(\frac{y}{e} \rightarrow 0\right)$

We get:

$$
=\lim _{y \rightarrow 0} \frac{\left(\frac{y}{e}\right)}{y}=\lim _{y \rightarrow 0} \frac{1}{e}=\frac{1}{e}
$$

## Problem Kuznetsov Limits 20-1