Question

Solve:
$\underset{x\to 0}{lim}\frac{\sqrt[3]{31+x}-\sqrt[3]{31-x}}{x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{{}^3\sqrt{1+x}{-}^3\sqrt{1-x}}{x}$

It is $\frac{0}{0}$ form, by using L, Hospital rule

$\begin{array}{c}{lim}_{x\to 0}\frac{{\left(1+x\right)}^{\frac{1}{3}}-{\left(1-x\right)}^{\frac{1}{3}}}{x}\\ {lim}_{x\to 0}\left\{\frac{1}{3}{\left(1+x\right)}^{\frac{-2}{3}}+\frac{1}{3}{\left(1-x\right)}^{\frac{-2}{3}}\right\}\\ =\frac{1}{3}\times 1+\frac{1}{3}\times 1=\frac{2}{3}.\end{array}$

Hence, this is the answer.