How to calculate $\lim_{x \to 0}\frac{\sqrt[3]{\cos x}-\sqrt{\cos x}}{x^2}$

Question

I found this exercise in my math book but there's no step by step solution for solving this without L'hospital rule. I have no idea how to start. $$\lim_{x \to 0}\frac{\sqrt[3]{\cos x}-\sqrt{\cos x}}{x^2}$$

Answer 1

Hint

$$a-b={a^6-b^6\over a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5}$$

substitute $a=\sqrt[3]{\cos x}$ and $b=\sqrt{\cos x}$ and then use L^Hopital's rule or Taylor expansion.

Answer 2

you can also use taylor series expansions. after differentiating $\sqrt[3]{\cos(x)}$ and $\sqrt{\cos(x)}$ a few times you'll get $$\sqrt[3]{\cos (x)}-\sqrt{\cos (x)}=\frac{x^2}{12}-\frac{x^4}{288}+O\left(x^5\right)$$ finding the limit is easy from there.