I'm trying to evaluate the following limit: $$\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$$ I've tried multiplying by the conjugate and variable substitution. I had a look at wolfram alpha and it said that $\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}=\sqrt{2}$, though I'm interested in the process to achieve that.

Any help would be much appreciated / actually finding the limit.


  • $\begingroup$ $cos(x)=1-\frac{x^2}{2}+o(x^3)$. $\endgroup$ – Qing Zhang Feb 18 '17 at 20:07
  • $\begingroup$ You can use L'Hopital. $\endgroup$ – Smurf Feb 18 '17 at 20:08

Note: I am using the limit $\lim_{\theta \to 0}\frac{\sin \theta}{\theta}=1$ and the identity $1-\cos 2A=2\sin^2 A$.

\begin{align*} \lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x} & = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 \left(x^2/2\right)}}{2 \sin^2 \left(x/2\right)}\\ & = \lim_{x \to 0} \frac{\sin \left(x^2/2\right)}{\sqrt{2}\sin^2 \left(x/2\right)}\\ & = \frac{1}{\sqrt{2}}\lim_{x \to 0} \frac{\sin \left(x^2/2\right)}{x^2/2}\frac{(x/2)^2}{\sin^2 \left(x/2\right)}.2\\ & =\sqrt{2}. \end{align*}



Using $$1-\cos(x)=\frac{x^2}{2}+o(x^2)$$ and $$1-\cos(x^2)=\frac{x^4}{2}+o(x^4),$$ you'll get the result.

  • $\begingroup$ What is the $o(x^2)$ mean? What is that called? $\endgroup$ – frog1944 Feb 18 '17 at 20:09
  • $\begingroup$ a function $f(x)=o(x)$ if $$\lim_{x\to 0}\left|\frac{f(x)}{x}\right|=0.$$ but to be honest, if you don't know Taylor approximation, my answer won't help. $\endgroup$ – Surb Feb 18 '17 at 20:11
  • $\begingroup$ Ok, thanks anyway $\endgroup$ – frog1944 Feb 18 '17 at 20:18

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