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

Question

Maybe you have any tips on evaluating this limit?

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

Answer 1

We have $$\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$$ for $\theta$ near $0$. Therefore,$$ \lim_{x\to0}\frac{\sqrt{1-\cos x^2}}{1-\cos x}=\lim_{x\to0}\frac{\sqrt 2\sin(x^2/2)}{1-\cos x} $$ and now you can use l'Hôpital's rule.

Answer 2

$$=\lim_{x\to0}\dfrac{\sin(x^2)}{\sin^2x}\cdot\dfrac{1+\cos x}{\sqrt{1+\cos( x^2)}}$$ as $\sin(x^2)>0$ for $x\to0$

Answer 3

Here's a tip:

$$ \cos( \theta ) = \cos( 2 \frac{\theta}{2} ) = \cos^2(\frac{\theta}{2}) - \sin^2(\frac{\theta}{2}) $$

$$ 1= \cos^2(\frac{\theta}{2}) + \sin^2(\frac{\theta}{2}) $$

$$ 1 - \cos( \theta ) = 2 \sin^2(\frac{\theta}{2}) $$

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