Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}$

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

$=\underset{x\to 0}{lim}\frac{(x+\frac{1}{6}{x}^3+\frac{3}{40}{x}^5+......)-(x-\frac{1}{3}{x}^3+\frac{2}{15}{x}^5-.........)}{x^3}$

$=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$