Functions Question 230
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{{{\sin }^{-1}}x-{{\tan }^{-1}}x}{x^{3}} $ is equal to
[RPET 2000]
Options:
A) 0
B) 1
C) ?1
D) $ 1/2 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 0}{\mathop{\lim }},\frac{{{\sin }^{-1}}x-{{\tan }^{-1}}x}{x^{3}} $ , $ ( \frac{0}{0} ) $ Applying L-Hospital?s rule, $ \underset{x\to 0}{\mathop{\lim }},\frac{\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{1+x^{2}}}{3x^{2}} $ , $ ( \frac{0}{0} ) $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{\frac{-1}{2}\times \frac{-2x}{{{(1-x^{2})}^{3/2}}}+\frac{2x}{{{(1+x^{2})}^{2}}}}{6x} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{1}{6}[ \frac{1}{{{(1-x^{2})}^{3/2}}}+\frac{2}{{{(1+x^{2})}^{2}}} ],=,\frac{1}{2} $ .
BETA
