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?

Remove the square root by observing that $1-\cos x\ge0$, so $1-\cos x=\sqrt{(1-\cos x)^2}$; hence you reduce to computing
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}
$$
and then take the square root of the result. Now
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}=
\lim_{x\to0}\frac{1-(1-x^4/2+o(x^4))}{(1-(1-x^2/2+o(x^2))^2}=
\lim_{x\to0}\frac{x^4/2+o(x^4)}{x^4/4+o(x^4)}=2
$$
You can even avoid using Taylor's expansion by recalling that
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}
$$
so you can directly write
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}=
\lim_{x\to0}\frac{1-\cos(x^2)}{x^4}\left(\frac{x^2}{1-\cos x}\right)^2=
\frac{1}{2}\cdot 4=2
$$
