Functions Question 270
Question: $ \underset{x\to -2}{\mathop{\lim }},\frac{{{\sin }^{-1}}(x+2)}{x^{2}+2x} $ is equal to
[Orissa JEE 2002]
Options:
A) 0
B) $ \infty $
C) ?1/2
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\underset{x\to -2}{\mathop{\lim }},\frac{{{\sin }^{-1}}(x+2)}{x^{2}+2x} $ , $ ( \frac{0}{0}form ) $ Using L-Hospital?s rule
Þ $ y=\underset{x\to -2}{\mathop{\lim }},\frac{( \frac{1}{\sqrt{1-{{(x+2)}^{2}}}} )}{2x+2} $
Þ $ y=\frac{1}{-4+2}=-\frac{1}{2} $ .
BETA
