#### Answer

The solution is
$$\lim_{x\to 1}\frac{4\tan^{-1}x-\pi}{x-1}=2.$$

#### Work Step by Step

We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to 1}\frac{4\tan^{-1}x-\pi}{x-1}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to 1}\frac{(4\tan^{-1}x-\pi)'}{(x-1)'}=\lim_{x\to 1}\frac{4\frac{1}{1+x^2}}{1}=\lim_{x\to 1}\frac{4}{1+x^2}=\left[\frac{4}{1+1^2}\right]=2.$$