<s>[INST] compute limit (no l'Hospital rule) I need to compute 
$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}.$$
I can not use the l'Hospital rule.
 [/INST] $$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{\cos x} -1+1- \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{1+\cos x-1} -1- (\sqrt[3]{1+\cos x-1}-1)}{\cos x-1}\cdot\frac{\cos x-1}{\sin^2x}=\lim_{x\to 0}[\frac{(1+\cos x-1)^\frac{1}{2}-1}{\cos x-1}-\frac{(1+\cos x-1)^\frac{1}{3}-1}{\cos x-1}]\cdot\frac{-1}{2}=(\frac{1}{2}-\frac{1}{3})\cdot\frac{-1}{2}=-\frac{1}{12}$$
We applied:$$ \lim_{t\to 0}\frac{(1+t)^r-1}{t} =r$$
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