## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 4} \frac{\sqrt{1+2 x}-3}{\sqrt{x}-2}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 4} \frac{\sqrt{1+2 x}-3}{\sqrt{x}-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 4} \frac{(\sqrt{1+2 x}-3)(\sqrt{1+2 x}+3)}{(\sqrt{x}-2)(\sqrt{1+2 x}+3)}= \\
& =\lim _{x \rightarrow 4} \frac{1+2 x-9}{(\sqrt{x}-2)(\sqrt{1+2 x}+3)}=\lim _{x \rightarrow 4} \frac{2(x-4)}{(\sqrt{x}-2)(\sqrt{1+2 x}+3)}= \\
& =\lim _{x \rightarrow 4} \frac{2(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{1+2 x}+3)}=\lim _{x \rightarrow 4} \frac{2 \sqrt{x}+4}{\sqrt{1+2 x}+3}= \\
& =\frac{2 \sqrt{4}+4}{\sqrt{1+2 \cdot 4}+3}=\frac{2 \cdot 2+4}{\sqrt{9}+3}=\frac{8}{6}=\frac{4}{3}=1 \frac{1}{3}
\end{aligned}
$$

## Problem Kuznetsov Limits 11-1