Calculating limits using l'Hôpital's rule. After a long page of solving limits using l'Hôpital's rule 
only those 2 left that i cant manage to solve
$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt[3]{\cos x}\over \sin^2 x }$$
$$\lim\limits_{x\to\ {pi\over 2}}{\tan 3x - 3\over \tan x - 3 }$$
Thanks in advance for any help :)
i edit the second one in mistake i entered $0$ insted of $\pi\over 2$

$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt[3]{\cos x}\over \sin^2 x }=\lim\limits_{x\to0}{-\frac{1}{2}\sin x(\cos x)^{-\frac{1}{2}} + \frac{1}{3}\sin x (\cos x)^{-\frac{1}{3}}\over 2\sin x \cos x }=\lim\limits_{x\to0}{-\frac{1}{2}(\cos x)^{-\frac{1}{2}} + \frac{1}{3} (\cos x)^{-\frac{1}{3}}\over 2 \cos x }=-\frac{1}{12}$$
For the second one, you can evaluate the limit directly by substituting in $x=0$.
