## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 8} \frac{\sqrt{9+2 x}-5}{\sqrt[3]{x}-2}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 8} \frac{\sqrt{9+2 x}-5}{\sqrt[3]{x}-2}=\lim _{x \rightarrow 8} \frac{(\sqrt{9+2 x}-5)(\sqrt{9+2 x}+5)}{(\sqrt[3]{x}-2)(\sqrt{9+2 x}+5)}= \\
& =\lim _{x \rightarrow 8} \frac{9+2 x-25}{(\sqrt[3]{x}-2)(\sqrt{9+2 x}+5)}=\lim _{x \rightarrow 8} \frac{2 x-16}{(\sqrt[3]{x}-2)(\sqrt{9+2 x}+5)}= \\
& =\lim _{x \rightarrow 8} \frac{(2 x-16)\left(\sqrt[3]{x^{2}}+2 \sqrt[3]{x}+4\right)}{(\sqrt[3]{x}-2)\left(\sqrt[3]{x^{2}}+2 \sqrt[3]{x}+4\right)(\sqrt{9+2 x}+5)}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{x \rightarrow 8} \frac{2(x-8)\left(\sqrt[3]{x^{2}}+2 \sqrt[3]{x}+4\right)}{(x-8)(\sqrt{9+2 x}+5)}=\lim _{x \rightarrow 8} \frac{2\left(\sqrt[3]{x^{2}}+2 \sqrt[3]{x}+4\right)}{\sqrt{9+2 x}+5}= \\
& =\frac{2\left(\sqrt[3]{8^{2}}+2 \sqrt[3]{8}+4\right)}{\sqrt{9+2 \cdot 8}+5}=\frac{2\left(2^{2}+2 \cdot 2+4\right)}{\sqrt{9+2 \cdot 8}+5}= \\
& =\frac{2(4+4+4)}{\sqrt{25}+5}=\frac{2 \cdot 12}{10}=\frac{12}{5}=2 \frac{2}{5}
\end{aligned}
$$

## Problem Kuznetsov Limits 11-26