<s>[INST] Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$ I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.
$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
Is there some other more effective way for this limit?
 [/INST] Well the idea is very simple. We have to use the fundamental limit $$\lim_{y \to 0}\frac{1 - \cos y}{y^{2}} = \frac{1}{2}$$ This is a pretty standard result which can easily be proved by simplifying $(1 - \cos y)$ as $2\sin^{2}(y/2)$ and then applying $\lim\limits_{y \to 0}\dfrac{\sin y}{y} = 1$.
For the current question we first put $\sqrt[3]{\cos x} = t$ so that $\cos x = t^{3}$ and then $$1 - \sqrt[3]{\cos x} = 1 - t = \dfrac{1 - t^{3}}{1 + t + t^{2}} = \frac{1 - \cos x}{1 + \sqrt[3]{\cos x} + \sqrt[3]{\cos^{2}x}}$$ We can now proceed as follows: $$\begin{aligned}L &= \lim_{x \to 0}\frac{1 - \sqrt[3]{\cos x}}{1 - \cos\sqrt[3]{x}}\\
&= \lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\cdot\frac{1}{1 + \sqrt[3]{\cos x} + \sqrt[3]{\cos^{2}x}}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{3}\cdot\frac{1}{2}\lim_{x \to 0}\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{6}\lim_{x \to 0}\dfrac{x^{2}}{\dfrac{1 - \cos\sqrt[3]{x}}{\sqrt[3]{x^{2}}}\cdot\sqrt[3]{x^{2}}}\\
&= \frac{1}{6}\lim_{x \to 0}x^{4/3}\cdot\frac{x^{2/3}}{1 - \cos (x^{1/3})}\\
&= \frac{1}{6}\cdot 0\cdot 2 = 0\end{aligned}$$
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