## Problem Statement

Calculate the limit of the function:

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

$$
\begin{aligned}
& \lim _{x \rightarrow 5} \frac{\sqrt[3]{9 x}-3}{\sqrt{3+x}-\sqrt{2 x}}=\lim _{x \rightarrow 3} \frac{(\sqrt[3]{9 x}-3)\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}{(\sqrt{3+x}-\sqrt{2 x})\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}= \\
& =\lim _{x \rightarrow 3} \frac{9 x-27}{(\sqrt{3+x}-\sqrt{2 x})\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}= \\
& =\lim _{x \rightarrow-3} \frac{(9 x-27)(\sqrt{3+x}+\sqrt{2 x})}{(\sqrt{3+x}-\sqrt{2 x})(\sqrt{3+x}+\sqrt{2 x})\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{x \rightarrow 3} \frac{(9 x-27)(\sqrt{3+x}+\sqrt{2 x})}{(3-x)\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}=\lim _{x \rightarrow 3} \frac{-9(3-x)(\sqrt{3+x}+\sqrt{2 x})}{(3-x)\left(\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9\right)}= \\
& =\lim _{x \rightarrow 3} \frac{-9(\sqrt{3+x}+\sqrt{2 x})}{\sqrt[3]{(9 x)^{2}}+\sqrt[3]{9 x} \cdot 3+9}=\frac{-9(\sqrt{3+3}+\sqrt{2 \cdot 3})}{\sqrt[3]{(9 \cdot 3)^{2}}+\sqrt[3]{9 \cdot 3} \cdot 3+9}= \\
& =\frac{-9(2 \sqrt{6})}{\sqrt[3]{27^{2}}+\sqrt[3]{27} \cdot 3+9}=\frac{-18 \sqrt{6}}{3^{2}+3 \cdot 3+9}=\frac{-18 \sqrt{6}}{27}=-\frac{2 \sqrt{6}}{3}
\end{aligned}
$$

## Problem Kuznetsov Limits 11-15