Question

Find the following limit:
$\underset{x\to -8}{lim}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{3x}}.$

Answer 1

$\underset{x\to -8}{lim}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{3x}}$

Using L hospital Rule

$=\underset{x\to -8}{lim}\frac{\frac{d}{dx}(\sqrt{1-x}-3)}{\frac{d}{dx}(2+\sqrt[3]{3x})}$

$\Rightarrow \underset{x\to -8}{lim}\frac{\frac{-1}{2\sqrt{1-x}}}{\frac{1}{3}{x}^{\frac{1}{3}-1}}$

$\Rightarrow \underset{x\to -8}{lim}\frac{-\frac{1}{2\times 3}}{\frac{1}{3}(-2{)}^{-2/3}}$

$=\frac{-\frac{1}{6}}{\frac{1}{3}\times \frac{1}{4}}=\frac{-4}{2}=-2$.