## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{3}+1}{\sin (x+1)}$
## Solution

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{y \rightarrow 0} \frac{y^{3}-3 y^{2}+3 y}{y}=\lim _{y \rightarrow 0} y^{2}-3 y+3=0^{2}-3 \cdot 0+3=3$

## Problem Kuznetsov Limits $16-3$