## Problem Statement

Calculate the limit of the function:

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

We will use 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:

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

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

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

$=\lim _{x \rightarrow 0}\left(1+\sqrt{x^{2}+1}\right)=\left(1+\sqrt{0^{2}+1}\right)=2$

## Problem Kuznetsov Limits 12-28