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  4. The value of displaystyle limx →0 (1/x) [ tan-1 ((x+1/2x+1))- (π/4)] is

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Q. The value of $\displaystyle\lim_{x \to0} \frac{1}{x} \left[\tan^{-1} \left(\frac{x+1}{2x+1}\right)- \frac{\pi}{4}\right] $ is

Limits and Derivatives

Solution:

$\displaystyle\lim_{x \to0} \frac{1}{x} \left[\tan^{-1} \left(\frac{x+1}{2x+1}\right)- \frac{\pi}{4}\right] $
$ = \displaystyle\lim _{x \to 0} \left(\frac{1}{x} \right) \left[ \tan^{-1} \left(\frac{x+1}{2x+1}\right) - \tan^{-1} \left(1\right) \right] $
$= \displaystyle\lim _{x \to 0} \left(\frac{1}{x} \right) . \tan ^{-1} \left(\frac{\frac{x+1}{2x+1} -1}{1+ \frac{x+1}{2x+1}}\right) $
$= \displaystyle\lim _{x \to 0} \frac{1}{x} . \tan^{-1} \left(\frac{-x}{3x+2}\right)$
$ =- \displaystyle\lim _{x \to 0} \left[\frac{\tan^{-1}\left(\frac{x}{3x+2}\right)}{\frac{x}{3x+2}} \times\frac{1}{3x+2}\right] =- \frac{1}{2} $