Question

Calculate the following limits.
$\underset{x\to 0}{lim}\frac{\sqrt[3]{31+\sin x}-\sqrt[3]{31-\sin x}}{x}.$

Answer 1

$\underset{x\to 0}{lim}\frac{\sqrt[3]{31+\sin x}-\sqrt[3]{31-\sin x}}{x}\left[\frac{0}{0}form\right]$

Apply L' hospital Rule.

$\Rightarrow \underset{x\to 0}{lim}\frac{\frac{\cos x}{3(1+\sin x{)}^{\frac{2}{3}}}-\frac{(-\cos x)}{3(1-\sin x{)}^{\frac{2}{3}}}}{1}$

$\underset{x\to 0}{lim}\frac{\cos x}{3}[\frac{1}{(1+\sin x{)}^{\frac{2}{3}}}+\frac{1}{(1-\sin x{)}^{\frac{2}{3}}}]$

Now put the limit,

$\Rightarrow \frac{1}{3}[1+1]=\frac{2}{3}[\because \cos 0=1$ and $\sin 0=0]$.