Answer
$$\lim_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{\tan^{-1}x}\Big)=0$$
Work Step by Step
$$A=\lim_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{\tan^{-1}x}\Big)$$
$$A=\lim_{x\to0^+}\frac{\tan^{-1}x-x}{x\tan^{-1}x}$$
$\lim_{x\to0^+}(\tan^{-1}x-x)=\tan^{-1}0-0=0$ and $\lim_{x\to0^+}(x\tan^{-1}x)=0\times\tan^{-1}0=0.$
For this being an indeterminate form of $\frac{0}{0}$, L'Hospital's Rule can be used:
$$A=\lim_{x\to0^+}\frac{(\tan^{-1}x-x)'}{(x\tan^{-1}x)'}$$
(Don't forget that $(\tan^{-1}x)'=\frac{1}{1+x^2}$)
$$A=\lim_{x\to0^+}\frac{\frac{1}{1+x^2}-1}{\tan^{-1}x+x\times\frac{1}{1+x^2}}$$
$$A=\lim_{x\to0^+}\frac{\frac{1-(1+x^2)}{1+x^2}}{\frac{\tan^{-1}x(1+x^2)+x}{1+x^2}}$$
$$A=\lim_{x\to0^+}\frac{\frac{-x^2}{1+x^2}}{\frac{\tan^{-1}x(1+x^2)+x}{1+x^2}}$$
$$A=\lim_{x\to0^+}\frac{-x^2}{\tan^{-1}x(1+x^2)+x}.$$
$\lim_{x\to0^+}(-x^2)=-0^2=0$ and $\lim_{x\to0^+}(\tan^{-1}x(1+x^2)+x)=\tan^{-1}0(1+0^2)+0=0\times1+0=0.$
Another indeterminate form of $\frac{0}{0}$. We use L'Hospital's Rule one more time:
$$A=\lim_{x\to0^+}\frac{(-x^2)'}{[\tan^{-1}x(1+x^2)+x]'}$$
$$A=\lim_{x\to0^+}\frac{-2x}{\frac{1}{1+x^2}(1+x^2)+\tan^{-1}x(2x)+1}$$
$$A=\lim_{x\to0^+}\frac{-2x}{1+\tan^{-1}x(2x)+1}$$
$$A=\lim_{x\to0^+}\frac{-2x}{2x\tan^{-1}x+2}$$
$$A=\frac{-2\times0}{2\times0\times\tan^{-1}0+2}$$
$$A=\frac{0}{0+2}=0$$
