Trigonometric Limit: $\lim_{x\to 0} \frac{\sqrt {\cos x} - \sqrt[3] {\cos x}}{\sin^2x}$ Compute the following limit:
$$\lim_{x\to 0} \frac{\sqrt {\cos x}  - \sqrt[3] {\cos x}}{\sin^2x}$$
How would I go about solving this, I can't used l´Hôpital

Writing $c$ and $s$ for $\cos x$ and $\sin x$ respectively,
$$\frac{c^{1/2} - c^{1/3}}{s^2} = \frac{c^{1/3}}{s^2} (c^{1/6} - 1)  = \frac{c -1}{s^2} \frac{c^{1/3}}{c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1} \ \ \ --(*) $$
as $(c^{1/6} - 1)(c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1) = c - 1$
Now $$c - 1 = \cos x - 1 = -2\sin^2(x/2) \ \ \hbox{ and} \ \ s^2 = \sin^2 x = 4\sin^2(x/2)\cos^2(x/2)$$ Hence the first fraction of (*)
$$ \frac{c -1}{s^2} = -\frac{1}{2}\cos^2(x/2) \longrightarrow -\frac{1}{2}, \ \ \hbox{ as } x \rightarrow 0$$
The second fraction $\displaystyle \frac{c^{1/3}}{c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1} \ \longrightarrow \frac{1}{6}, \ \ \hbox{ as } x \rightarrow 0$
Hence $$\lim_{x\rightarrow 0} \frac{c^{1/2} - c^{1/3}}{s^2} = -\frac{1}{2} . \frac{1}{6} = -\frac{1}{12}$$
