## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}$
## Solution

$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\lim _{x \rightarrow 0} \frac{1}{x+6}}=$

$=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{0+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{6}}=$

Using the substitution of equivalent infinitesimals:

$\sin 5 x^{2} \sim 5 x^{2}$, as $x \rightarrow 0\left(5 x^{2} \rightarrow 0\right)$
$\sin x \sim x$, as $x \rightarrow 0(x \rightarrow 0)$

We get:

$=\left(\lim _{x \rightarrow 0} \frac{5 x^{2}}{x}\right)^{\frac{1}{6}}=\left(\lim _{x \rightarrow 0} 5 x\right)^{\frac{1}{6}}=0^{\frac{1}{6}}=0$

## Problem Kuznetsov Limits 18-26