Answer
$0$
Work Step by Step
Given
$$\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{\tan ^{-1} x}\right)$$
Simplify and apply L'Hopital's rule
\begin{aligned}
\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{\tan ^{-1} x}\right)&=
\lim _{x \rightarrow 0^{+}}\left(\frac{\tan^{-1}x-x}{x\tan ^{-1} x}\right) \\
&= \lim _{x\to \:0+}\left(\frac{\frac{1}{x^2+1}-1}{\tan^{-1} \left(x\right)+\frac{x}{x^2+1}}\right),\ \ \ \text{Apply L'Hopital's rule }\\
&= \lim _{x\to \:0+}\left(\frac{-\frac{2x}{\left(x^2+1\right)^2}}{\frac{2}{\left(x^2+1\right)^2}}\right),\ \ \ \text{Apply L'Hopital's rule }\\
&= \frac{-\frac{2\cdot \:0}{\left(0^2+1\right)^2}}{\frac{2}{\left(0^2+1\right)^2}}\\
&= 0
\end{aligned}
