Question

$\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^3}$=

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

Answer 1

The correct option is D $\frac{1}{2}$

We know ${\tan }^{-1}x=x-\frac{1}{3}{x}^3+......$
and ${\sin }^{-1}x=x+\frac{1}{6}{x}^3+......$
$\therefore \underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^3}=\underset{x\to 0}{lim}\frac{(x+\frac{x^3}{6}+.....)-(x-\frac{x^3}{3}+..)}{x^3}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$