<s>[INST] The limit of $(1-\cos^{1/3} x)/(1-\cos x^{1/3})$ as $x\to 0$ I have a problem to find answer for this limit.
$$\lim_{x\to 0}\frac{1-{\cos^{1/3} x}}{1-\cos(x^{1/3})}$$
I did it like this $$\dfrac{\dfrac{\sin x}{3×\sqrt[3]{\cos x²}}}{\sin \sqrt[3]x×\dfrac 1{3×\sqrt[3]{x²}}}$$
 [/INST] We need two standard limits $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{1 - \cos^{1/3}x}{1 - \cos x^{1/3}}\notag\\
&= \lim_{x \to 0}\frac{1 - \cos^{1/3}x}{1 - \cos x}\cdot\frac{1 - \cos x}{1 - \cos x^{1/3}}\notag\\
&= \lim_{t \to 1}\frac{1 - t^{1/3}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos x}{1 - \cos x^{1/3}}\text{ (putting }t = \cos x)\notag\\
&= \frac{1}{3}\cdot 1^{-2/3}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{x^{2}}{1 - \cos x^{1/3}}\notag\\
&= \frac{1}{6}\cdot\lim_{x \to 0}x^{4/3}\cdot\frac{x^{2/3}}{1 - \cos x^{1/3}}\notag\\
&= \frac{1}{6}\cdot 0\cdot 2 = 0\notag
\end{align}
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