A) 2
B) \[\frac{1}{2}\]
C) \[-\frac{1}{3}\]
D) None of these
Correct Answer: B
Solution :
[b] \[\underset{x\to \infty }{\mathop{\lim }}\,x\left[ {{\tan }^{-1}}\left( \frac{x+1}{x+2} \right)-{{\tan }^{-1}}\left( \frac{x}{x+2} \right) \right]\] \[=\underset{x\to \infty }{\mathop{Lim}}\,x{{\tan }^{-1}}\left( \frac{\frac{x+1}{x+2}-\frac{x}{x+2}}{1+\frac{x+1}{x+2}.\frac{x}{x+2}} \right)\] \[=\underset{x\to \infty }{\mathop{Lim}}\,x{{\tan }^{-1}}\left( \frac{x+2}{2{{x}^{2}}+5x+4} \right)\] \[=\underset{x\to \infty }{\mathop{Lim}}\,x\left( \frac{{{\tan }^{-1}}\left( \frac{x+2}{2{{x}^{2}}+5x+4} \right)}{\frac{x+2}{2{{x}^{2}}+5x+4}} \right)\times \frac{x(x+2)}{2{{x}^{2}}+5x+4}\] \[=1\times \frac{1}{2}=\frac{1}{2}\]
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