<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] By L'Hopital's rule, we have
$$\begin{aligned}
\lim_{x \to 0}\frac{1 - \cos(x^2)}{(1 - \cos(x))^2}
&= \lim_{x \to 0}\frac{2x\sin(x^2)}{2(1 - \cos(x))\sin(x)} \\
&= \lim_{x \to 0}\frac{x}{\sin(x)}\frac{\sin(x^2)}{1 - \cos(x)} \\
&= \lim_{x \to 0}\frac{\sin(x^2)}{1 - \cos(x)} \\
\end{aligned}$$
where we have used that 
$$\lim_{x \to 0}\frac{x}{\sin(x)} = 1$$
Now we can apply L'Hopital's rule again to obtain
$$\begin{aligned}
\lim_{x \to 0}\frac{\sin(x^2)}{1 - \cos(x)}
&= \lim_{x \to 0} \frac{2x\cos(x^2)}{\sin(x)} \\
&= \lim_{x \to 0} \frac{x}{\sin(x)} 2\cos(x^2) \\
&= \lim_{x \to 0} 2\cos(x^2) \\
&= 2
\end{aligned}$$
Summarizing what we have so far,
$$\lim_{x \to 0} \frac{1 - \cos(x^2)}{(1 - \cos(x))^2} = 2$$
Now take the square root of both sides and use the fact that $\sqrt{(\cdot)}$ is continuous at zero (note that we approach only from the right) to conclude that
$$
\begin{aligned}
\lim_{x \to 0}\frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)}
&= \lim_{x \to 0}\sqrt{\frac{1 - \cos(x^2)}{(1 - \cos(x))^2}} \\
&= \sqrt{\lim_{x \to 0} \frac{1 - \cos(x^2)}{(1 - \cos(x))^2} } \\
&= \sqrt{2}
\end{aligned}$$
as desired.
</s>