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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 $x=0$.
As $\displaystyle\sqrt{\cos x}-\sqrt[3]{\cos x}=(\cos x)^{\frac12}-(\cos x)^{\frac13}$ and gcd$\displaystyle\left(\frac12,\frac13\right)=\frac{\text{gcd}(1,1)}{\text{lcm}(2,3)}=\frac16,$
Set $\displaystyle\sqrt[6]{\cos x}=1+h\implies\cos x=(1+h)^6$
and $\displaystyle\sqrt{\cos x}-\sqrt[3]{\cos x}=(1+h)^3-(1+h)^2=(1+h)^2(1+h-1)$
Finally, $\displaystyle\sin^2x=1-(1+h)^{12}=-12h+O(h^2)$
@copper.hat thanks i think i got it
$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt[3]{\cos x}\over \sin^2 x } = \lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt[3]{\cos x}\over 1-\cos^2 x } = \lim\limits_{y\to1}{\sqrt {y} - \sqrt[3]{y}\over 1- y^2 } $$ $$\lim\limits_{y\to1}{\sqrt {y} - \sqrt[3]{y}\over 1- y^2 } = \lim\limits_{y\to1}{\sqrt {y} - \sqrt[3]{y}\over 2y(6\sqrt[6] y) } = 0$$
