## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 3} \frac{\ln (2 x-5)}{e^{\sin \pi x}-1}
$$
## Solution

Substitution:

$x=y+3 \Rightarrow y=x-3$

$x \rightarrow 3 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{\ln (2 x-5)}{e^{\sin \pi x}-1}=\lim _{y \rightarrow 0} \frac{\ln (2(y+3)-5)}{e^{\sin \pi(y+3)}-1}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (2 y+6-5)}{e^{\sin (\pi y+3 \pi)}-1}=\lim _{y \rightarrow 0} \frac{\ln (2 y+1)}{e^{\sin (\pi y+\pi)}-1}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (2 y+1)}{e^{-\sin \pi y}-1}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_c071f2ede8d7a98b06e8g-09.jpg?height=73&width=822&top_left_y=1809&top_left_x=150)

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

We get:

$$
=\lim _{y \rightarrow 0} \frac{2 y}{-\sin \pi y}=
$$

Using the substitution of equivalent infinitesimals:

$\sin \pi y \sim \pi y$, as $y \rightarrow 0(\pi y \rightarrow 0)$

We get:

$$
=\lim _{y \rightarrow 0} \frac{2 y}{-\pi y}=\lim _{y \rightarrow 0} \frac{2}{-\pi}=-\frac{2}{\pi}
$$

## Problem Kuznetsov Limits 14-17