## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 4} \frac{\sqrt[3]{16 x}-4}{\sqrt{4+x}-\sqrt{2 x}}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 4} \frac{\sqrt[3]{16 x}-4}{\sqrt{4+x}-\sqrt{2 x}}=\lim _{x \rightarrow 4} \frac{(\sqrt[3]{16 x}-4)\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}{(\sqrt{4+x}-\sqrt{2 x})\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}= \\
& =\lim _{x \rightarrow 4} \frac{16 x-64}{(\sqrt{4+x}-\sqrt{2 x})\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}= \\
& =\lim _{x \rightarrow 4} \frac{(16 x-64)(\sqrt{4+x}+\sqrt{2 x})}{(\sqrt{4+x}-\sqrt{2 x})(\sqrt{4+x}+\sqrt{2 x})\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}= \\
& =\lim _{x \rightarrow 4} \frac{(16 x-64)(\sqrt{4+x}+\sqrt{2 x})}{(4-x)\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}=\lim _{x \rightarrow 4} \frac{-16(4-x)(\sqrt{4+x}+\sqrt{2 x})}{(4-x)\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}= \\
& =\lim _{x \rightarrow 4} \frac{-16(\sqrt{4+x}+\sqrt{2 x})}{\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16}=\frac{-16(\sqrt{4+4}+\sqrt{2 \cdot 4})}{\sqrt[3]{(16 \cdot 4)^{2}}+\sqrt[3]{16 \cdot 4} \cdot 4+16}= \\
& =\frac{-16(2 \sqrt{8})}{\sqrt[3]{64}+\sqrt[3]{64} \cdot 4+16}=\frac{-16 \cdot 4 \sqrt{2}}{4^{2}+4 \cdot 4+16}=\frac{-16 \cdot 4 \sqrt{2}}{16 \cdot 3}=-\frac{4 \sqrt{2}}{3}
\end{aligned}
$$

## Problem Kuznetsov Limits 11-17