## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow-8} \frac{10-x-6 \sqrt{1-x}}{2+\sqrt[3]{x}}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow-8} \frac{10-x-6 \sqrt{1-x}}{2+\sqrt[3]{x}}=\lim _{x \rightarrow-8} \frac{(10-x-6 \sqrt{1-x})(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})}= \\
& =\lim _{x \rightarrow-8} \frac{(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{100-20 x+x^{2}-36+36 x}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}= \\
& =\lim _{x \rightarrow-8} \frac{64+16 x+x^{2}}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{(8+x)^{2}}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{x \rightarrow-8} \frac{\left((2+\sqrt[3]{x})\left(4-2 \sqrt[3]{x}+\sqrt[3]{x^{2}}\right)\right)^{2}}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{(2+\sqrt[3]{x})\left(4-2 \sqrt[3]{x}+\sqrt[3]{x^{2}}\right)^{2}}{10-x+6 \sqrt{1-x}}= \\
& =\frac{(2+\sqrt[3]{-8})\left(4-2 \sqrt[3]{-8}+\sqrt[3]{(-8)^{2}}\right)^{2}}{10-(-8)+6 \sqrt{1-(-8)}}=\frac{(2-2)\left(4-2 \cdot(-2)+(-2)^{2}\right)^{2}}{10+8+6 \sqrt{9}}= \\
& =\frac{0(4+4+4)^{2}}{10+8+6 \sqrt{9}}=0
\end{aligned}
$$

## Problem Kuznetsov Limits 11-30