Question

${lim}_{x\to 0}\frac{\sqrt[3]{31+sinx}-\sqrt[3]{31-sinx}}{x}$

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

Answer 1

Answer: (A)

$\frac{2}{3}$

$\underset{x\to 0}{lim}\frac{\sqrt[3]{31+\sin x}-\sqrt[3]{31-\sin x}}{x}$
Applying L'Hospital rule
$=\underset{x\to 0}{lim}\frac{\frac{\cos x}{3{(1-\sin x)}^{2/3}}+\frac{\cos x}{3{(1+\sin x)}^{2/3}}}{1}$
$=\frac{1}{3}(\underset{x\to 0}{lim}\cos x{(1-\sin x)}^{-2/3}+\underset{x\to 0}{lim}\cos x{(1+\sin x)}^{-2/3})$
$=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$