<s>[INST] 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.
 [/INST] 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
$$\frac{432+\pi ^2}{216 \sqrt{2}}\approx 1.44652$$
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