Mathematics - Limits Question with Solution | TestHub

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