3
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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

$$\frac{1-\sqrt[3]{\cos x}}{1-\cos \sqrt[3]{x}} = \frac{\frac{x^2}{6} + O(x^4)}{\frac{x^{2/3}}{2} + O(x^{4/3})} = \frac{x^{4/3}}{3} + O(x^2) \to 0.$$

$\endgroup$
3
$\begingroup$

Yes there's the Taylor series:

$$\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}\sim_0 \frac{1-\left(1-\frac{x^2}{2}\right)^{1/3}}{1-\left(1-\frac{(x^{1/3})^2}{2}\right)}\sim_0\frac{\frac{x^2}{6}}{\frac{x^{2/3}}{2}}=\frac13 x^{4/3}\xrightarrow{x\to0}0$$

$\endgroup$
3
$\begingroup$

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: $$\begin{aligned}L &= \lim_{x \to 0}\frac{1 - \sqrt[3]{\cos x}}{1 - \cos\sqrt[3]{x}}\\ &= \lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\cdot\frac{1}{1 + \sqrt[3]{\cos x} + \sqrt[3]{\cos^{2}x}}\\ &= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\\ &= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\ &= \frac{1}{3}\cdot\frac{1}{2}\lim_{x \to 0}\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\ &= \frac{1}{6}\lim_{x \to 0}\dfrac{x^{2}}{\dfrac{1 - \cos\sqrt[3]{x}}{\sqrt[3]{x^{2}}}\cdot\sqrt[3]{x^{2}}}\\ &= \frac{1}{6}\lim_{x \to 0}x^{4/3}\cdot\frac{x^{2/3}}{1 - \cos (x^{1/3})}\\ &= \frac{1}{6}\cdot 0\cdot 2 = 0\end{aligned}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.