#### Answer

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

#### Work Step by Step

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