A) 2
B) 1/3
C) 2/3
D) -1/3
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{lim}}\,\,\,f(x)=\underset{x\to 0}{\mathop{lim}}\,\left( \frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x} \right)=f(0),\,\,\left( \frac{0}{0}\,\,form \right)\] Applying L-Hospital?s rule \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 2-\frac{1}{\sqrt{1-{{x}^{2}}}} \right)}{\left( 2+\frac{1}{1+{{x}^{2}}} \right)}\,\,=\frac{2-1}{2+1}=\frac{1}{3}\]You need to login to perform this action.
You will be redirected in
3 sec