## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}$
## Solution

$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\lim _{x \rightarrow 0} \frac{3}{x+8}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{0+8}}=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{8}}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1+x^{2}\right) \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$

We get:

$=\left(\lim _{x \rightarrow 0} \frac{x^{2}}{x^{2}}\right)^{\frac{3}{8}}=\left(\lim _{x \rightarrow 0} 1\right)^{\frac{3}{8}}=1^{\frac{3}{8}}=1$

## Problem Kuznetsov Limits 18-21