A) 1
B) \[-\frac{1}{2}\]
C) 2
D) 0
Correct Answer: B
Solution :
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\left\{ {{\tan }^{-1}}\left( \frac{x+1}{2x+1} \right)-\frac{\pi }{4} \right\}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\left\{ {{\tan }^{-1}}\left( \frac{x+1}{2x+1} \right)-{{\tan }^{-1}}(1) \right\}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\cdot {{\tan }^{-1}}\left\{ \frac{\frac{x+1}{2x+1}-1}{1+\frac{x+1}{2x+1}} \right\}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\cdot {{\tan }^{-1}}\left\{ \frac{x+1-2x-1}{2x+1+x+1} \right\}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\cdot {{\tan }^{-1}}\left\{ \frac{-x}{4x+2} \right\}\to \] from\[\,\left( \frac{0}{0} \right)\] Use L-Hospital rule \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{1+\frac{{{x}^{2}}}{{{(4x+2)}^{2}}}}\times \left[ -\left( \frac{4x+2-4x}{{{(4x+2)}^{2}}} \right) \right]\] \[=\underset{x\to 0}{\mathop{\lim }}\,-\frac{(+2)}{{{x}^{2}}+{{(4x+2)}^{2}}}=-\frac{(+2)}{0+{{(0+2)}^{2}}}=\frac{-2}{4}=\frac{-1}{2}\]
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