A) 2
B) \[1/3\]
C) \[2/3\]
D) \[-1/3\]
Correct Answer: B
Solution :
\[f(x)=\underset{x\to 0}{\mathop{\lim }}\,\,\left( \frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x} \right)=f(0)\], \[\left( \frac{0}{0} \right)\] Applying L-Hospital?s rule, \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 2-\frac{1}{\sqrt{1-{{x}^{2}}}} \right)}{\left( 2+\frac{1}{1+{{x}^{2}}} \right)}\] \[=\frac{2-1}{2+1}=\frac{1}{3}\] Trick : f (0) =\[\underset{x\to 0}{\mathop{\lim }}\,\frac{2-\frac{{{\sin }^{-1}}x}{x}}{2+\frac{{{\tan }^{-1}}x}{x}}=\frac{2-1}{2+1}=\frac{1}{3}\].
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