Question

Find the value of the limit: ${lim}_{x\to 0}\frac{{\sin }^{-1}x-\sin x}{x^3}$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

$\frac{1}{3}$

${lim}_{x\to 0}\frac{si{n}^{-1}x-sinx}{x^3}=\frac{0}{0}$ Form.

By L-Hospital Rule

${lim}_{x\to 0}\frac{\frac{1}{\sqrt{-x^2+1}}-cosx}{3x^2}=\frac{0}{0}$ Form.

${lim}_{x\to 0}\frac{\frac{x}{(1-x^2{)}^{3/2}+sinx}}{6x}=\frac{0}{0}$ Form.

again L-Hospital.

${lim}_{x\to 0}\frac{(1-x^2{)}^{3/2}+x.\frac{3}{2}(1-x^2{)}^{1/2}.2x+cosx}{6}=\frac{2}{6}$

$=\frac{1}{3}$