A) \[0\]
B) \[2\]
C) \[1/2\]
D) \[1\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{\sin }^{-1}}\,(x-2)}{{{x}^{2}}-4}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{\sin }^{-1}}(x-2)}{(x-2)\,(x+2)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{\sin }^{-1}}(x-2)}{x-2}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x+2}\] \[=1\times \frac{1}{0+2}\] \[\left( \because \,\,\underset{\theta \to 0}{\mathop{\lim }}\,\,\,\frac{{{\sin }^{-1}}\theta }{\theta }=1 \right)\] \[=\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
