Question

$\underset{x\to 0}{Lim}\frac{\sqrt[3]{31+\sin x}-\sqrt[3]{31-\sin x}}{x}=$

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

Answer 1

Answer: (C)

$\frac{2}{3}$

Now,

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

Now applying L'Hospital's rule we get,

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

$=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}$.