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!
 A: So, this is asking, by definition of derivative, $\arctan'(\sqrt3)$. That is $\displaystyle\frac1{1+3}$.
A: We want
$$L = \lim_{x \to \sqrt{3}} \dfrac{\arctan(x) - \pi/3}{x - \sqrt3}$$
Let $\arctan(x) = t$. We then have
$$L = \lim_{t \to \pi/3} \dfrac{t-\pi/3}{\tan(t) - \sqrt{3}} = \lim_{t \to \pi/3} \dfrac{t-\pi/3}{\tan(t) - \tan(\pi/3)} = \dfrac1{\left.\dfrac{d \tan(t)}{dt} \right\vert_{t=\pi/3}} = \dfrac1{\sec^2(t) \vert_{t=\pi/3}} = \dfrac14$$
A: Putting $$\frac\pi3-\tan^{-1}x=\theta\implies \tan^{-1}x=\frac\pi3-\theta$$
So, $$x=\tan\left(\frac\pi3-\theta\right)\text{ and }x-\sqrt3=\tan\left(\frac\pi3-\theta\right)-\tan\frac\pi3=\frac{\sin(\frac\pi3-\theta-\frac\pi3)}{\cos\frac\pi3\cos \left(\frac\pi3-\theta\right)}=-\frac{2\sin\theta}{\cos\left(\frac\pi3-\theta\right)}$$
$$\text{So,}\lim_{x\rightarrow\sqrt{3}} \frac{\tan^{-1} x - \frac{\pi}{3}}{x-\sqrt{3}}=\lim_{\theta\to0}\frac{-\theta}{-\frac{2\sin\theta}{\cos\left(\frac\pi3-\theta\right)}}=\frac12 \cdot \frac{\lim_{\theta\to0}\cos\left(\frac\pi3-\theta\right)}{\lim_{\theta\to0}\frac{\sin\theta}\theta}=\frac{\cos\frac\pi3}{2\cdot1}=\frac1{2\cdot2}$$
