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I don't understand the rewriting that's being done in this limit:

$$\lim_{x\to0} \frac{1−\cos x}{\sin x} = \lim_{x\to0} \frac{\sin x}{\cos x} $$

Why doesn't this simplify to $\frac{\sin x}{\sin x}$?

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    $\begingroup$ L'Hopitals Rule. $\endgroup$ – ireallydonknow Nov 28 '14 at 11:43
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    $\begingroup$ and $\sin x/\sin x = 1$ whereas the original limit is $0$ $\endgroup$ – Ilya Nov 28 '14 at 11:43
  • $\begingroup$ Welcome to Math.SE! Please take a look at math.stackexchange.com/help/notation to see how to use MathJax on this site. $\endgroup$ – AlexR Nov 28 '14 at 11:47
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It doesn't. You can use l'Hospital to get $$\lim_{x\to0} \frac{1-\cos x}{\sin x} = \lim_{x\to0} \frac{\sin x}{\cos x} = \tan 0 = 0$$

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rewrite it as $$\frac{1-\cos(x)}{\sin(x)}\frac{1+\cos(x)}{1+\cos(x)}$$

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  • $\begingroup$ It's a good idea to find the value of the limit, but it boils down to $\frac{\sin x}{1+\cos x}$ which is something that the OP got $\endgroup$ – Ilya Nov 28 '14 at 11:51
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$$\frac{1-\cos x}{\sin x}=\frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}=\tan(x/2).$$

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Another idea is $$\frac{1-\cos x}{\sin x} = \frac{1-\cos x}{x}\cdot\frac{x}{\sin x}=\frac{1-\cos x}{x}\cdot\frac{1}{\frac{\sin x}{x}}$$

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