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I need to evaluate :

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

In terms of the number:

$\alpha = \lim_{x\to0} \frac{\sin(x)}{x}$

What I am thinking is that $\cos(x) = \frac{\sin(2x)}{2\sin(x)}$

But I don't know if this is the correct beginning

One example (I am doing this just to show that I can't derivate) I already did was:

$\lim_{x\to0} \frac{\sin(ax)}{\sin(bx)} =\lim_{x\to0} \frac{\sin(ax)\cdot bx\cdot a}{ax\cdot \sin(bx)\cdot b} = \alpha \cdot \frac{1}{\alpha}\cdot\frac{a}{b} = \frac{a}{b}$

Winther
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Juju9708
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    You could just apply L’Hospital’s rule (if allowed) – WW1 Apr 22 '18 at 19:32
  • Yes WW1, but as I stated on my example, I am not allowed to derivate, with L'Hopital it would be 1/2 I guess, but that doesn't work for me. – Juju9708 Apr 22 '18 at 19:33
  • You could try to use that $[1-\cos(x)][1+\cos(x)] = \sin^2(x)$ to show that the limit is equal the limit of $\frac{1}{1+\cos(x)}\left(\frac{\sin(x)}{x}\right)^2$ – Winther Apr 22 '18 at 19:40

1 Answers1

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Hint: $$ 1-\cos x=2\sin^2\frac{x}{2}\Rightarrow \lim_{x\to0} \frac{1-\cos(x)}{x^2}=\lim_{x\to0}\frac{2\sin^2\frac{x}{2}}{x^2}=? $$

user
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