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 $\pi\over 2$