Solve the limit $\lim\limits _{x\to 0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}$ $$\lim _{x\to \:0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}=\left|\frac{0}{0}\right|$$
I think you have to multiply by the conjugate. And then make the change equivalent small. Right?

I am waiting for a simpler more succinct answer but in the meantime we can use Taylor's theorem
$$ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{4n} = 1 - \frac{x^4}{2} + \frac{x^8}{120} + O(x^{12})$$
$$ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2} + \frac{x^4}{120} + O(x^{6}) $$
Now we have
$$\lim_{x \to 0} \frac{\sqrt{\frac{x^4}{2} - \frac{x^8}{120} + O(x^{12})}}{\frac{x^2}{2} - \frac{x^4}{120} + O(x^{6})} = \lim_{x \to 0} \frac{\sqrt{\frac{1}{2} - \frac{x^4}{120} + O(x^{8})}}{\frac{1}{2} - \frac{x^2}{120} + O(x^{4})} = \frac{\sqrt{\frac{1}{2}}}{\frac{1}{2}}$$
$$ \lim_{x \to 0} \frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)} = \sqrt{2}$$
I don't really like using Taylor's theorem for "simple" limits, so I await a cleverer answer!
