## Problem Statement

Calculate the limit of the function:

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

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+x^{2}}-(1+x)}{x}=\left\{\frac{0}{0}\right\}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+x^{2}-(1+x)^{2}}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}=\lim _{x \rightarrow 0} \frac{1-2 x+x^{2}-\left(1^{2}+2 x+x^{2}\right)}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+x^{2}-1-2 x-x^{2}}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}=\lim _{x \rightarrow 0} \frac{-4 x^{2}}{x\left(\sqrt{1-2 x+x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{-4}{\sqrt{1-2 x+x^{2}}+(1+x)}=\frac{-4}{\sqrt{1-2 \cdot 0+0^{2}}+(1+0)}= \\
& =\frac{-4}{\sqrt{1}+1}=\frac{-4}{2}=-2
\end{aligned}
$$

## Problem Kuznetsov Limits 11-8