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The value of $\lim\limits_{n\to \infty}x\bigg[\tan^{-1}\large\frac{x+1}{x+2}$$-\cot^{-1}\large\frac{x+2}{x}\bigg]$ is

$(a)\;-1\qquad(b)\;\large\frac{-1}{2}$$\qquad(c)\;\large\frac{1}{2}$$\qquad(d)\;1$

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1 Answer

$\lim\limits_{x\to \infty}x\bigg[\tan^{-1}\large\frac{x+1}{x+2}$$-\tan^{-1}\large\frac{x}{x+2}\bigg]$
$\Rightarrow \lim\limits_{x\to \infty}x\tan^{-1}\bigg[\large\frac{\Large\frac{x+1}{x+2}-\frac{x}{x+2}}{1+\Large\frac{x+1}{x+2}.\frac{x}{x+2}}\bigg]$
$\Rightarrow \lim\limits_{x\to \infty}x\tan^{-1}\bigg[\large\frac{x+2}{2x^2+5x+4}\bigg]$
$\lim\limits_{x\to \infty}\large\frac{\tan^{-1}\Large\frac{x+2}{2x^2+5x+4}}{\Large\frac{x+2}{2x^2+5x+4}}\times \large\frac{x(x+2)}{2x^2+5x+4}$
$\Rightarrow 1\times \large\frac{1}{2}$
$\Rightarrow \large\frac{1}{2}$
Hence (c) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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