Limit $\frac{\tan^{-1}x - \tan^{-1}\sqrt{3}}{x-\sqrt{3}}$ without L'Hopital's rule. Please solve this without L'Hopital's rule? $$\lim_{x\rightarrow\sqrt{3}} \frac{\tan^{-1} x - \frac{\pi}{3}}{x-\sqrt{3}}$$
All I figured out how to do is to rewrite this as $$\frac{\tan^{-1} x - \tan^{-1}\sqrt{3}}{x-\sqrt{3}}$$
Any help is appreciated!

We want
$$L = \lim_{x \to \sqrt{3}} \dfrac{\arctan(x) - \pi/3}{x - \sqrt3}$$
Let $\arctan(x) = t$. We then have
$$L = \lim_{t \to \pi/3} \dfrac{t-\pi/3}{\tan(t) - \sqrt{3}} = \lim_{t \to \pi/3} \dfrac{t-\pi/3}{\tan(t) - \tan(\pi/3)} = \dfrac1{\left.\dfrac{d \tan(t)}{dt} \right\vert_{t=\pi/3}} = \dfrac1{\sec^2(t) \vert_{t=\pi/3}} = \dfrac14$$
