Question

The value of $\underset{x\to 0}{lim}\frac{{(1+x)}^{1/3}-{(1-x)}^{1/3}}{x}is$

• A. $\frac{1}{2}$
• B. $0$
• C. $-1$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

We have,

${lim}_{x\to 0}\frac{{(1+x)}^{\frac{1}{3}}-{(1-x)}^{\frac{1}{3}}}{x}$

Applying L’ Hospital rule and we get,

${lim}_{x\to 0}\frac{\frac{1}{3}{(1+x)}^{\frac{1}{3}-1}-\frac{1}{3}{(1-x)}^{\frac{1}{3}-1}(-1)}{1}$

$\Rightarrow {lim}_{x\to 0}\frac{1}{3}{(1+x)}^{\frac{1}{3}-1}-\frac{1}{3}{(1-x)}^{\frac{1}{3}-1}(-1)$

Taking limit and we get,

$=\frac{1}{3}{(1+0)}^{\frac{1}{3}-1}+\frac{1}{3}{(1-0)}^{\frac{1}{3}-1}$

$=\frac{2}{3}$

Hence, this is the answer.