Question

$\lim_{x \rightarrow 0}\frac{\sin^{- 1}x - \tan^{- 1}x}{x^{3}}$ is equal to

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

Answer 1

Answer: (D)

$\frac{1}{2}$

Answer 2

$\lim_{x \rightarrow 0}\frac{\sin^{- 1}x - \tan^{- 1}x}{x^{3}}$ $\left( \frac{0}{0}\text{form} \right)$

Applying L-Hospital’s rule,

$= \lim_{x \rightarrow 0}\frac{\frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{1 + x^{2}}}{3x^{2}}$ $\left( \frac{0}{0}\text{form} \right)$

$= \lim_{x \rightarrow 0}\frac{\frac{- 1}{2} \times \frac{- 2x}{(1 - x^{2})^{3/2}} + \frac{2x}{(1 + x^{2})^{2}}}{6x} = \lim_{x \rightarrow 0}\frac{1}{6}\left\lbrack \frac{1}{(1 - x^{2})^{3/2}} + \frac{2}{(1 + x^{2})^{2}} \right\rbrack = \frac{1}{2}$.