A) \[-\frac{1}{3}\]
B) \[1\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: D
Solution :
Given, \[f(x)=\left\{ \begin{matrix} \frac{2x-si{{n}^{-1}}x}{2x+{{\tan }^{-1}}x}, & x\ne 0 \\ k\,(let), & x=0 \\ \end{matrix} \right.\] Since, \[f(x)\] is continuous at \[x=0.\] \[\therefore \] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=f(0)\] ?..(i) \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(0-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{-2h-{{\sin }^{-1}}(-h)}{-2h+{{\tan }^{-1}}(-h)}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{-2+\frac{1}{\sqrt{1-{{h}^{^{2}}}}}}{-2-\frac{1}{1+{{h}^{2}}}}\] \[=\frac{-2+\frac{1}{\sqrt{1-0}}}{-2-\frac{1}{1+0}}\] \[=\frac{-1}{-3}=\frac{1}{3}\] From Eq. (i), \[f(0)=\frac{1}{3}\]
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