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How does one evaluate the limit

$$\lim_{x\to 0} \frac{\sqrt{1-\cos{(x^2)}}}{1-\cos{x}}$$

Without using series expansion or L'hopital's rule? I know it's pretty straightforward to use a series expansion to get the limit of $\sqrt{2}$ but I wonder whether it is possible to evaluate this without such methods.

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  • $\begingroup$ Edited the title as I put limits to infinity rather than 0. $\endgroup$ – Loo Soo Yong Mar 10 '18 at 8:04
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For $0<|x|<1$, $$\begin{align}\frac{\sqrt{1-\cos(x^2)}}{1-\cos x}&=\frac{\sqrt{1-\left(1-2\sin^2\left(\frac{x^2}2\right)\right)}}{1-\left(1-2\sin^2\left(\frac x2\right)\right)}=\frac{\sqrt2\sin\left(\frac{x^2}2\right)}{2\sin^2\left(\frac x2\right)}\\ &=\left(\sqrt2\right)\left(\frac{\sin\left(\frac{x^2}2\right)}{\frac{x^2}2}\right)\left(\frac{\frac x2}{\sin\left(\frac x2\right)}\right)^2\end{align}$$ So $$\lim_{x\rightarrow0}\frac{\sqrt{1-\cos(x^2)}}{1-\cos x}=\left(\sqrt2\right)\left(1\right)\left(1\right)^2=\sqrt2$$

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Note that by standard limit

$$\frac{\sqrt{1-\cos(x^2)}}{1-\cos x}=\frac{\sqrt{1-\cos(x^2)}}{x^2}\frac{x^2}{1-\cos x}=\sqrt{\frac{1-\cos(x^2)}{x^4}}\frac{x^2}{1-\cos x}\to\frac1{\sqrt 2}\frac1{\frac12}=\sqrt 2$$

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