## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 3} \frac{\sqrt[3]{5+x}-2}{\sin \pi x}$
## Solution

Substitution:

$$
\begin{aligned}
& x=y+3 \Rightarrow y=x-3 \\
& x \rightarrow 3 \Rightarrow y \rightarrow 0
\end{aligned}
$$

## We get:

$\lim _{x \rightarrow 3} \frac{\sqrt[3]{5+x}-2}{\sin \pi x}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{5+(y+3)}-2}{\sin \pi(y+3)}=$

$=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{\sin (\pi y+3 \pi)}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{\sin (\pi y+\pi)}=$

$=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{-\sin \pi y}=$

Using the substitution of equivalent infinitesimals:

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

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\sqrt[3]{y+8}-2}{-\pi y}= \\
& =\lim _{y \rightarrow 0} \frac{(\sqrt[3]{y+8}-2)\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}{-\pi y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y+8-8}{-\pi y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y}{-\pi y\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\
& =\lim _{y \rightarrow 0} \frac{1}{-\pi\left(\sqrt[3]{(y+8)^{2}}+2 \sqrt[3]{y+8}+4\right)}= \\
& =\frac{1}{-\pi\left(\sqrt[3]{(0+8)^{2}}+2 \sqrt[3]{0+8}+4\right)}=\frac{1}{-\pi\left(2^{2}+2 \cdot 2+4\right)}= \\
& =-\frac{1}{12 \pi}
\end{aligned}
$$

Problem Kuznetsov Limits 16-23