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?
 A: Taylor expansions yield the result without much difficulty. We have
$$\sqrt[3]{1+y} = 1 + \frac{y}{3} + O(y^2),$$
and
$$\cos z = 1 - \frac{z^2}{2} + O(z^4),$$
hence
$$\sqrt[3]{\cos x} = \sqrt[3]{1 - x^2/2 + O(x^4)} = 1- \frac{x^2}{6} + O(x^4),$$
so
$$\frac{1-\sqrt[3]{\cos x}}{1-\cos \sqrt[3]{x}} = \frac{\frac{x^2}{6} + O(x^4)}{\frac{x^{2/3}}{2} + O(x^{4/3})} = \frac{x^{4/3}}{3} + O(x^2) \to 0.$$
A: 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: $$\begin{aligned}L &= \lim_{x \to 0}\frac{1 - \sqrt[3]{\cos x}}{1 - \cos\sqrt[3]{x}}\\
&= \lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\cdot\frac{1}{1 + \sqrt[3]{\cos x} + \sqrt[3]{\cos^{2}x}}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{3}\cdot\frac{1}{2}\lim_{x \to 0}\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{6}\lim_{x \to 0}\dfrac{x^{2}}{\dfrac{1 - \cos\sqrt[3]{x}}{\sqrt[3]{x^{2}}}\cdot\sqrt[3]{x^{2}}}\\
&= \frac{1}{6}\lim_{x \to 0}x^{4/3}\cdot\frac{x^{2/3}}{1 - \cos (x^{1/3})}\\
&= \frac{1}{6}\cdot 0\cdot 2 = 0\end{aligned}$$
A: Yes there's the Taylor series:
$$\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}\sim_0 \frac{1-\left(1-\frac{x^2}{2}\right)^{1/3}}{1-\left(1-\frac{(x^{1/3})^2}{2}\right)}\sim_0\frac{\frac{x^2}{6}}{\frac{x^{2/3}}{2}}=\frac13 x^{4/3}\xrightarrow{x\to0}0$$
A: You can simply use L'Hospital's Rule $$\lim_{x\to 0}\frac{1-(\cos x)^\frac13}{1-\cos (x)^\frac13}\space=\space\lim_{x\to 0}\frac{\frac{(\cos 0)^\frac13-(\cos x)^\frac13}{0-x}}{\frac{\cos (0)^\frac13-\cos(x)^\frac13}{0-x}}\space=\space\frac{\lim_{x\to 0}\frac{(\cos x)^\frac13-(\cos 0)^\frac13}{x-0}}{\lim_{x\to 0}\frac{\cos(x)^\frac13-\cos(0)^\frac13}{x-0}}$$ If you set $(\cos x)^\frac13$ and $\cos(x)^\frac13$ equal to $f(x)$ and $g(x)$ respectively, you will get $$\lim_{x\to 0}\frac{f'(x)}{g'(x)}\space=\space\lim_{x\to 0}\frac{\frac{(\cos x)^\frac{-2}{3}(-\sin x)}{3}}{\sin(x)^\frac{1}{3}x^\frac{-2}{3}}\space=\space\lim_{x\to 0}\frac{\sin x (x^\frac23)}{3\sin(x)^\frac13 (\cos x)^\frac23}$$ One more time, divide both denominator and nominator by $x$ $$\lim_{x\to 0}\frac{\frac{\sin x (x)^\frac23}{x}}{\frac{3\sin(x)^\frac13 (\cos x)^\frac23}{(x^\frac13)(x^\frac23)}}\space=\space\lim_{x\to 0}\frac{x^\frac43}{3(\cos x)^\frac23}\space=\space 0$$
A: Writing $x=u^3$ for convenience, we have
$$\begin{align}
{1-\sqrt[3]{\cos x}\over1-\cos\sqrt[3]x}
&={1-\sqrt[3]{\cos u^3}\over1-\cos u}\\
&={1-\cos u^3\over1-\cos u}\cdot{1\over1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3}}\\
&={1-\cos^2u^3\over1-\cos^2u}\cdot{1+\cos u\over(1+\cos u^3)(1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3})}\\
&=u^4\left(\sin u^3\over u^3\right)^2\left(u\over\sin u\right)^2{1+\cos u\over(1+\cos u^3)(1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3})}\\
&\to0\cdot1\cdot1\cdot{1+1\over(1+1)(1+1+1)}\\
&=0
\end{align}$$
