calculus - Evaluating $lim_{x to 0} frac{sqrt{1- cos x^2}}{1 - cos x}$

I'm trying to evaluate the following limit:
$$\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$$
I've tried multiplying by the conjugate and variable substitution. I had a look at wolfram alpha and it said that $\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}=\sqrt{2}$, though I'm interested in the process to achieve that.

Any help would be much appreciated / actually finding the limit.

Thanks

Answer

Note: I am using the limit $\lim_{\theta \to 0}\frac{\sin \theta}{\theta}=1$ and the identity $1-\cos 2A=2\sin^2 A$.

$$\begin{align*} \lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x} & = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 \left(x^2/2\right)}}{2 \sin^2 \left(x/2\right)}\\ & = \lim_{x \to 0} \frac{\sin \left(x^2/2\right)}{\sqrt{2}\sin^2 \left(x/2\right)}\\ & = \frac{1}{\sqrt{2}}\lim_{x \to 0} \frac{\sin \left(x^2/2\right)}{x^2/2}\frac{(x/2)^2}{\sin^2 \left(x/2\right)}.2\\ & =\sqrt{2}. \end{align*}$$