limits of inverse trigonometric functions without L'hospital's rule How do I solve this without using L'Hospital's rule? $$\lim_{h\rightarrow0} \frac{\cos^{-1}(\frac{1}{2}-h) -\cos^{-1}(\frac{1}{2})}{h}$$
I already tried letting $\theta=\cos^{-1}(\frac{1}{2}-h)$ gets $\cos\theta=\frac{1}{2}-h$ then $h=\frac{1}{2}-\cos\theta$, replaced all $h$ with this and I'm lost. I think this doesn't help. I'm getting a zero as value or indefinite one which shouldn't be because the value must be $\frac{2\sqrt{3}}{3}$. Please help.
 A: Your start is good.
$$\lim_{h\to0}\frac{\cos^{-1}\left(\frac{1}{2}-h\right)-\cos^{-1}\left(\frac{1}{2}\right)}{h}$$
Let $\theta=\cos^{-1}\left(\frac{1}{2}-h\right)$ so $h=\frac{1}{2}-\cos h$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\frac{1}{2}-\cos\theta}$$
Rewrite fraction as a trig value and apply sums to products.
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\cos\left(\frac{\pi}{3}\right)-\cos\theta}$$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{2\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\frac{\theta}{2}-\frac{\pi}{6}}{\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\frac{\theta}{2}-\frac{\pi}{6}}{\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)}\times\lim_{\theta\to\frac{\pi}{3}}\frac{1}{\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$
$$=1\times\frac{1}{\frac{\sqrt{3}}{2}}$$
$$=\frac{2}{\sqrt{3}}$$
$$=\frac{2\sqrt{3}}{3}$$
