The value of $\underset{x\to 0}{lim} \frac{1}{x}\left[{\tan }^{-1}\left(\frac{x+1}{2x+1}\right)-\frac{\pi }{4}\right]$ is

$\text{ Limit }\pi /4={\tan }^{-1}1$ Use 

${\tan }^{-1}x-{\tan }^{-1}y=\tan \frac{(x-y)}{1+xy}$
$$\begin{gathered} \underset{x\to 0}{lim} \frac{1}{x}\left[{\tan }^{-1}\left(\frac{x+1}{2x+1}\right)-\frac{\pi }{4}\right] \\ =\underset{x\to 0}{lim} \frac{1}{x}\left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{x+1}{2x+1}-1}}{\frac{x+1}{2x+1}+1}\right)\right] \\ =\underset{x\to 0}{lim} \frac{1}{x}{\tan }^{-1}\left(\frac{-x}{3x+2}\right) \\ =-\underset{x\to 0}{lim}\frac{{\tan }^{-1}\left(\frac{x}{3x+2}\right)}{\frac{x}{3x+2}\left(3x+2\right)}=-\frac{1}{2} \end{gathered}$$