## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 3} \frac{\sqrt{x+13}-2 \sqrt{x+1}}{\sqrt[3]{x^{2}-9}}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{\sqrt{x+13}-2 \sqrt{x+1}}{\sqrt[3]{x^{2}-9}}=\lim _{x \rightarrow 3} \frac{(\sqrt{x+13}-2 \sqrt{x+1})(\sqrt{x+13}+2 \sqrt{x+1})}{\sqrt[3]{x^{2}-9}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{x+13-4(x+1)}{\sqrt[3]{x^{2}-9}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{-3 x+9}{\sqrt[3]{(x-3)(x+3)}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{-3(x-3)}{\sqrt[3]{x-3} \sqrt[3]{x+3}(\sqrt{x+13}+2 \sqrt{x+1})}= \\
& =\lim _{x \rightarrow 3} \frac{-3 \sqrt[3]{(x-3)^{2}}}{\sqrt[3]{x+3}(\sqrt{x+13}+2 \sqrt{x+1})}=\frac{-3 \sqrt[3]{(3-3)^{2}}}{\sqrt[3]{3+3}(\sqrt{3+13}+2 \sqrt{3+1})}= \\
& =\frac{-3 \sqrt[3]{0^{2}}}{\sqrt[3]{6}(\sqrt{16}+2 \sqrt{4})}=0
\end{aligned}
$$

## Problem Kuznetsov Limits 11-31