Differentiation Question 25
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{1}{x}{{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) $ is equal to
Options:
A) 1
B) 0
C) 2
D) none of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] we know that $ Co{s^{-1}}( \frac{1-x^{2}}{1+x^{2}} )= \begin{cases} 2{{\tan }^{-1}}x,x\ge 0 \\ -2{{\tan }^{-1}}x,x\le 0 \\ \end{cases} . $
Or $ \underset{x\to {0^{+}}}{\mathop{\lim }}\frac{1}{x}{{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} )=\underset{x\to {0^{+}}}{\mathop{\lim }}\frac{2{{\tan }^{-1}}x}{x}=2 $
And $ \underset{x\to {0^{-}}}{\mathop{\lim }}\frac{1}{x}{{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} )=\underset{x\to {0^{+}}}{\mathop{\lim }}[ -\frac{2{{\tan }^{-1}}x}{x} ]=-2 $
BETA
