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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Sign up to join this communityI 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 $\frac{\sin x}{\sin x}$?
It doesn't. You can use l'Hospital to get $$\lim_{x\to0} \frac{1-\cos x}{\sin x} = \lim_{x\to0} \frac{\sin x}{\cos x} = \tan 0 = 0$$
rewrite it as $$\frac{1-\cos(x)}{\sin(x)}\frac{1+\cos(x)}{1+\cos(x)}$$
$$\frac{1-\cos x}{\sin x}=\frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}=\tan(x/2).$$
Another idea is $$\frac{1-\cos x}{\sin x} = \frac{1-\cos x}{x}\cdot\frac{x}{\sin x}=\frac{1-\cos x}{x}\cdot\frac{1}{\frac{\sin x}{x}}$$