## Problem Statement

Calculate the limit of the function:

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

Substitution:

$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

$\lim _{x \rightarrow 1} \frac{e^{x}-e}{\sin \left(x^{2}-1\right)}=\lim _{y \rightarrow 0} \frac{e^{y+1}-e}{\sin \left((y+1)^{2}-1\right)}=$
$=\lim _{y \rightarrow 0} \frac{e\left(e^{y}-1\right)}{\sin \left(y^{2}+2 y+1-1\right)}=\lim _{y \rightarrow 0} \frac{e\left(e^{y}-1\right)}{\sin \left(y^{2}+2 y\right)}=$

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& e^{y}-1 \sim{ }_{,} \text {as } y \rightarrow 0 \\
& \sin \left(y^{2}+2 y\right) \sim y^{2}+2{ }_{, \text {as }} y \rightarrow 0\left(y^{2}+2 y \rightarrow 0\right)
\end{aligned}
$$

We get:

$=\lim _{y \rightarrow 0} \frac{e \cdot y}{y^{2}+2 y}=\lim _{y \rightarrow 0} \frac{e}{y+2}=\frac{e}{0+2}=\frac{e}{2}$

## Problem Kuznetsov Limits 16-31