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 $$\dfrac{\dfrac{\sin x}{3×\sqrt[3]{\cos x²}}}{\sin \sqrt[3]x×\dfrac 1{3×\sqrt[3]{x²}}}$$
 A: 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}
A: First of all, multiply and divide by
$$
1+\cos^{1/3}x+\cos^{2/3}x
$$
so that the limit becomes
$$
\lim_{x\to0}\frac{1-\cos x}{1-\cos x^{1/3}}
  \frac{1}{1+\cos^{1/3}x+\cos^{2/3}x}
$$
and the second fraction has limit $1/3$, so it can be pulled out, leaving to compute
$$
\lim_{x\to0}\frac{1-\cos x}{1-\cos x^{1/3}}=
\lim_{x\to0}\frac{1-\cos x}{x^2}\frac{x^{2/3}}{1-\cos x^{1/3}}x^{4/3}
$$
A: Applying l'Hopital's rule:
$$\lim_{x\to 0} \frac{1-\sqrt[3]{\cos(x)}}{1-\cos\left(\sqrt[3]{x}\right)}=$$
$$\lim_{x\to 0} \frac{\frac{\text{d}}{\text{d}x}\left(1-\sqrt[3]{\cos(x)}\right)}{\frac{\text{d}}{\text{d}x}\left(1-\cos\left(\sqrt[3]{x}\right)\right)}=$$
$$\lim_{x\to 0} \frac{\frac{\sin(x)}{3\cos^{\frac{2}{3}}(x)}}{\frac{\sin\left(\sqrt[3]{x}\right)}{3x^{\frac{2}{3}}}}=$$
$$\lim_{x\to 0} \frac{x^{\frac{2}{3}}\cdot\cos^{-\frac{2}{3}}(x)\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}=$$
$$\left(\lim_{x\to 0}\cos^{-\frac{2}{3}}(x)\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(\lim_{x\to 0}\frac{1}{\cos^{\frac{2}{3}}(x)}\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(\frac{1}{\cos^{\frac{2}{3}}(0)}\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(\frac{1}{1}\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(1\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}$$
Now l'Hopital's rule again, and you'll find the answer
A: Hint: Approximate both instances of $\cos(t)$ using the first two terms of its Taylor series, 
and then employ the binomial series approximation for $\sqrt[n]{1+u}\simeq1+\dfrac un$.
