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.
 A: if you multiply and divide by both conjugates you get:
$$\lim_{x \to 0} \frac{(1+\cos{x})\sin{(x^2)}}{\sqrt{1+\cos{(x^2)}} \space \sin^2{x}}$$
Amplifying to get limit of the form $\frac{\sin{x}}{x} :$
$$=\lim_{x \to 0} \frac{(1+\cos{x})}{\sqrt{1+\cos{x^2}}} \space \frac{\sin{(x^2)}}{x^2} \space \frac{x^2}{\sin^2{x}}=\frac{2}{\sqrt{2}}*1*1^2=\sqrt{2}$$
A: 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.
A: Just note that $1-\cos t=2\sin^2(t/2)$ and hence the expression under limit is equal to $$\frac{\sqrt{2}\sin(x^2/2)}{2\sin^2(x/2)}$$ which can be rewritten as $$\frac{1}{\sqrt{2}}\cdot\frac{\sin(x^2/2)}{x^2/2}\cdot \frac {x^2}{2}\cdot\frac{4}{x^2}\cdot\frac{(x/2)^2}{\sin^2(x/2)}$$ Using limit $\lim_{t\to 0}(\sin t) /t=1$ we get the desired limit as $$\frac{1}{\sqrt{2}}\cdot 1\cdot 2\cdot 1^2=\sqrt{2}$$
A: 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
$$\frac{432+\pi ^2}{216 \sqrt{2}}\approx 1.44652$$
A: An approach based on the fact that
$$
\lim_{u\to0} \frac{1-\cos u}{u^2} = \frac{1}{2} \tag{1}
$$
which is a standard limit (and equivalent to a Taylor expansion of $\cos$ to order $2$ at $0$).
$$
\frac{\sqrt{1-\cos(x^2)}}{1-\cos x}
= \sqrt{\frac{1-\cos(x^2)}{x^4}}\cdot\frac{x^2}{1-\cos x} \xrightarrow[x\to0]{}\sqrt{\frac{1}{2}}\cdot\frac{1}{\frac{1}{2}} = \sqrt{2}\,,
$$
applying (1) to $u=x^2$ and $u=x$ separately (for the first, since $x^2\to 0$ when $x\to 0$).

To be completely precise: we used (1) twice, and the continuity of both  $\sqrt{\cdot}$ and the inverse function at $1/2$ to have $$\lim_{x\to 0} \sqrt{g(x)}\frac{1}{h(x)} = \lim_{x\to 0}\sqrt{g(x)}\lim_{x\to 0}\frac{1}{h(x)} = \sqrt{\lim_{x\to 0}g(x)}\frac{1}{\lim_{x\to 0}h(x)}$$
A: \begin{align*}
\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}
&=\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}\cdot(1+\cos x)}{1-\cos^2(x)}
=2\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{\sin^2(x)}\\[10pt]
&=2\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{x^2}\cdot\frac{x^2}{\sin^2(x)}
=2\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{x^2}
\stackrel{y:=x^2}{=}2\lim_{y\to 0^+}\frac{\sqrt{1-\cos(y)}}{y}\\[10pt]
&=2\lim_{y\to 0^+}\sqrt{\frac{1-\cos(y)}2}\cdot\frac{\sqrt 2}{y}
=2\lim_{y\to 0^2}\sin\left(\frac y2\right)\cdot\frac{\sqrt 2}{y}
=\sqrt 2\lim_{y\to 0^+}\sin\left(\frac y2\right)\cdot\frac{2}{y}=\boldsymbol{\sqrt 2}
\end{align*}
