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The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\frac{\pi}{4}\right]$ is :
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Verified Answer
The correct answer is:
$-\frac{1}{2}$
$-\frac{1}{2}$
$\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\frac{\pi}{4}\right]$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\tan ^{-1}(1)\right]$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right) \cdot \tan ^{-1}\left(\frac{\frac{x+1}{2 x+1}-1}{1+\frac{x+1}{2 x+1}}\right)$
$=\lim _{x \rightarrow 0} \frac{1}{x} \cdot \tan ^{-1}\left(\frac{-x}{3 x+2}\right)$
$=-\lim _{x \rightarrow 0}\left[\frac{\tan ^{-1}\left(\frac{x}{3 x+2}\right)}{\frac{x}{3 x+2}} \times \frac{1}{3 x+2}\right]=-\frac{1}{2}$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\tan ^{-1}(1)\right]$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right) \cdot \tan ^{-1}\left(\frac{\frac{x+1}{2 x+1}-1}{1+\frac{x+1}{2 x+1}}\right)$
$=\lim _{x \rightarrow 0} \frac{1}{x} \cdot \tan ^{-1}\left(\frac{-x}{3 x+2}\right)$
$=-\lim _{x \rightarrow 0}\left[\frac{\tan ^{-1}\left(\frac{x}{3 x+2}\right)}{\frac{x}{3 x+2}} \times \frac{1}{3 x+2}\right]=-\frac{1}{2}$
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