Any idea how to find $\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$? $$\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $\sqrt{2}$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $\cos$.
Then i tried L'Hopital because it is $\frac{0}{0}$ and still that $\cos$ in the square root is doing problems. I tried on symbolab  calculator but it  can't solve it.
So can someone help me how to solve this?
Thank you.

If you want to go beyond the limit it self, use Taylor series and binomial expansion
$$\cos(x^2)=1-\frac{x^4}{2}+\frac{x^8}{24}+O\left(x^{12}\right)$$
$$1-\cos(x^2)=\frac{x^4}{2}-\frac{x^8}{24}+O\left(x^{12}\right)$$
$$\sqrt{1-\cos \left(x^2\right)}=\frac{x^2}{\sqrt{2}}-\frac{x^6}{24 \sqrt{2}}+O\left(x^{10}\right)$$
$$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)$$
$$\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\frac{\frac{x^2}{\sqrt{2}}-\frac{x^6}{24 \sqrt{2}}+O\left(x^{10}\right) }{ \frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)}$$ Now, long division to get
$$\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\sqrt{2}+\frac{x^2}{6 \sqrt{2}}+O\left(x^4\right)$$ which shows the limit and how it is approached.
For the fun, use your pocket calculator for $x=\frac \pi 6$. The exact value would be
$$2 \sqrt{2} \left(2+\sqrt{3}\right) \sin \left(\frac{\pi ^2}{72}\right)\approx 1.44244$$ while the above truncated series would give
$$\frac{432+\pi ^2}{216 \sqrt{2}}\approx 1.44652$$
