<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] As lcm $(2,3)=6, $ let $(\cos x)^\frac16=y$
$\displaystyle\implies (\cos x)^\frac12=y^3,(\cos x)^\frac13=y^2$ and $\displaystyle \cos x=y^6\implies\sin^2x=1-y^{12}$
$$\lim_{x\to0}\frac{(\cos x)^\frac12-(\cos x)^\frac13}{\sin^2x}=\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}\frac{-y^2(1-y)}{(1-y)(1+y+\cdots+y^{10}+y^{11})}$$
We can cancel $1-y$ as $1-y\ne0$ as $y\to1$

Alternatively, $$\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}(-y^2)\cdot\frac1{\lim_{y\to1}\frac{y^{12}-1}{y-1}}$$
Again, $$\lim_{y\to1}\frac{y^{12}-1}{y-1}=\lim_{y\to1}\frac{y^{12}-1^{12}}{y-1}=\frac{d(y^{12})}{dy}_{(\text{ at }y=1)}$$
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