Limit of: $\lim_{x\to0}\frac{1-\sqrt{\cos{x}}}{1-\cos{\sqrt{x}}}$

I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.

$\lim_{x\to0}\frac{1-\sqrt{\cos{x}}}{1-\cos{\sqrt{x}}}$

Is there some other more effective way for this limit?

Taylor expansions yield the result without much difficulty. We have

$$\sqrt{1+y} = 1 + \frac{y}{3} + O(y^2),$$

and

$$\cos z = 1 - \frac{z^2}{2} + O(z^4),$$

hence

$$\sqrt{\cos x} = \sqrt{1 - x^2/2 + O(x^4)} = 1- \frac{x^2}{6} + O(x^4),$$

so

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

Yes there's the Taylor series:

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

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 $\sqrt{\cos x} = t$ so that $\cos x = t^{3}$ and then $$1 - \sqrt{\cos x} = 1 - t = \dfrac{1 - t^{3}}{1 + t + t^{2}} = \frac{1 - \cos x}{1 + \sqrt{\cos x} + \sqrt{\cos^{2}x}}$$ We can now proceed as follows: \begin{aligned}L &= \lim_{x \to 0}\frac{1 - \sqrt{\cos x}}{1 - \cos\sqrt{x}}\\ &= \lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt{x}}\cdot\frac{1}{1 + \sqrt{\cos x} + \sqrt{\cos^{2}x}}\\ &= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt{x}}\\ &= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{x^{2}}{1 - \cos\sqrt{x}}\\ &= \frac{1}{3}\cdot\frac{1}{2}\lim_{x \to 0}\frac{x^{2}}{1 - \cos\sqrt{x}}\\ &= \frac{1}{6}\lim_{x \to 0}\dfrac{x^{2}}{\dfrac{1 - \cos\sqrt{x}}{\sqrt{x^{2}}}\cdot\sqrt{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}