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$$\lim _{x\to \:0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}=\left|\frac{0}{0}\right|$$

I think you have to multiply by the conjugate. And then make the change equivalent small. Right?

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No L'Hospital. No Taylor:

$$\lim_{x\to0}\frac{\sqrt{1-\cos x^2}}{1-\cos x}\\=\lim_{x\to0}\frac{\sqrt{2\sin^2(x^2/2)}}{2\sin^2(x/2)}\\=\lim_{x\to0}\frac{\sin(x^2/2)}{\sqrt{2}\sin^2(x/2)}\\=\lim_{x\to0}\underbrace{\frac{\sin(x^2/2)}{x^2/2}}_1\underbrace{\left(\frac{x/2}{\sin(x/2)}\right)^2}_{1^2}\frac{2}{\sqrt2}=\sqrt2$$

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  • $\begingroup$ Thank you for your response to me is most appropriate :) $\endgroup$ – andre1 Apr 11 '15 at 10:36
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One may recall that, as $x \to 0$, we have (by the Taylor expansion) $$ \cos x =1-\frac {x^2}{2}+\mathcal{O}(x^3), \quad \sqrt {1+x } =1+\mathcal{O}(x), $$ giving $$\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}=\frac{\sqrt{x^4/2+\mathcal{O}(x^6)}}{x^2/2+\mathcal{O}(x^3)}=\sqrt{2}\:\frac{\sqrt{1+\mathcal{O}(x)}}{1+\mathcal{O}(x)}$$ and the desired limit is $\sqrt{2} $.

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My approach would be to Taylor expand:

$$ \frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)} = \frac{\sqrt{1 - (1 - x^4/2)}}{1 - (1 - x^2/2)} + \mathcal{O}(x^3) = \frac{x^2/\sqrt{2}}{x^2/2} + \mathcal{O}(x^3)= \sqrt{2} + \mathcal{O}(x^3) $$

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Apply l'Hospital's Rule:

$$\lim_{x\to 0}\frac{\sqrt{1-\cos x^2}}{1-\cos x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{x\sin x^2}{\sin x\sqrt{1-\cos x^2}}=\lim_{x\to 0}\frac x{\sin x}\cdot\frac{\sin x^2}{\sqrt{1-\cos x^2}}$$

The first factor's limit is $\; 1\;$ , whereas the second factor gives:

$$\lim_{x\to 0}\frac{\sin x^2}{\sqrt{1-\cos x^2}}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{2\sqrt{1-\cos x^2}\cos x^2}{\sin x^2}=\lim_{x\to 0}2\cos x^2\frac{\sqrt{1-\cos x^2}}{\sin x^2}$$

$$=2\cdot\frac1{\sqrt2}=\sqrt2$$

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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 $$ \lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}= \lim_{x\to0}\frac{1-\cos(x^2)}{x^4}\left(\frac{x^2}{1-\cos x}\right)^2= \frac{1}{2}\cdot 4=2 $$

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Using finite expansions:

$$\lim_{x \to 0} \frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)} = \lim_{x \to 0} \frac{\sqrt{1 - (1 - \frac{(x^2)^2}{2!})}}{1 - (1 - \frac{x^2}{2!})} = \lim_{x \to 0} \frac{x^2}{x^2} \times \sqrt2 = \sqrt2$$

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  • $\begingroup$ "Finite expansions"? That's just Taylor up to the quadratic term, isn't it? $\endgroup$ – Timbuc Apr 11 '15 at 10:33
  • $\begingroup$ Thanks for the feedback guys. @Timbuc, I agree, we can't write $\cos(x) = 1 - \frac{x^2}{2!}$. But doesn't the presence of $\lim$ allow that writing? I mean, we are saying that the two expressions approach the same value near $0$, not that they have the same value near $0$. $\endgroup$ – goldenratio Apr 11 '15 at 11:02
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    $\begingroup$ @goldenratio Well, yes: the presence of that lim there really makes things correct... but ... I'd expect a little expanation there as the one you just provided in your comment, or at least leave this task as a hintfor the asker. Good point, indeed. $\endgroup$ – Timbuc Apr 11 '15 at 11:04
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I am waiting for a simpler more succinct answer but in the meantime we can use Taylor's theorem $$ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{4n} = 1 - \frac{x^4}{2} + \frac{x^8}{120} + O(x^{12})$$ $$ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2} + \frac{x^4}{120} + O(x^{6}) $$ Now we have $$\lim_{x \to 0} \frac{\sqrt{\frac{x^4}{2} - \frac{x^8}{120} + O(x^{12})}}{\frac{x^2}{2} - \frac{x^4}{120} + O(x^{6})} = \lim_{x \to 0} \frac{\sqrt{\frac{1}{2} - \frac{x^4}{120} + O(x^{8})}}{\frac{1}{2} - \frac{x^2}{120} + O(x^{4})} = \frac{\sqrt{\frac{1}{2}}}{\frac{1}{2}}$$ $$ \lim_{x \to 0} \frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)} = \sqrt{2}$$ I don't really like using Taylor's theorem for "simple" limits, so I await a cleverer answer!

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I have a way which only use the usual limit $$ \lim_{y\rightarrow 0}\frac{1-\cos y}{y^{2}}=\frac{1}{2}. $$

That is $$ \lim_{x\rightarrow 0}\frac{\sqrt{1-\cos (x^{2})}}{1-\cos x}% =\lim_{x\rightarrow 0}\sqrt{\frac{1-\cos (x^{2})}{(x^{2})^{2}}}\times \lim_{x\rightarrow 0}\frac{x^{2}}{1-\cos x}=\sqrt{\frac{1}{2}}\times 2=\sqrt{% 2}. $$

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