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I am working on a textbook problem. The first step is to prove that

$$\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$$

(which I did). The exercise goes on

Use this limit [i.e. the one above] to find $$\lim \limits_{x \to 0} \frac{1 - \cos x}{x}$$.

How can this be done? I don't really see a connection between the two...

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Use $\cos \theta=1-2\sin^2(\theta/2)$ – David Mitra Nov 13 '11 at 21:27
Hint: $1-\cos x=\frac{1-\cos^2 x}{1+\cos x}=\frac{\sin^2 x}{1+\cos x}$. – André Nicolas Nov 13 '11 at 21:28
@AndréNicolas: Maybe you should point that this is natural: $1-\cos(x)= 1- \sqrt{1-\sin^2(x)}$ and you rationalize.... – N. S. Nov 13 '11 at 21:44
@N.S.: A good comment for the OP to consider! I was just giving a hint on how to start. – André Nicolas Nov 13 '11 at 21:57
More solutions to this limit are here and here. – Lord_Farin Jun 15 '13 at 7:55
up vote 11 down vote accepted

Using half-angle formula, we have $1-\cos x = 2\sin^2\frac x2$ so $$ \lim\limits_{x\to 0}\frac{1-\cos x}{x} = 2\lim\limits_{x\to 0}\frac{\sin^2\frac x2}{x} = \lim\limits_{x\to 0}\frac{\sin\frac x2}{x/2}\cdot\lim\limits_{x\to 0}\sin\frac x2 = 1\cdot0 = 0 $$ where we used the fact that both limits exist.

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$$ \begin{eqnarray*} \lim_{x \to 0} \frac{1 - \cos{x}}{x} &=& \lim_{x \to 0} \frac{(1-\cos{x})(1+\cos x)}{x(1+\cos x)} \\ &=& \lim_{x \to 0} \frac{1-\cos^2 x}{x(1 + \cos x)} \\ &=& \lim_{x \to 0} \frac{x\sin^2 x}{x \cdot x(1+ \cos x)} \\ &=& \lim_{x \to 0} \frac{\sin x}{x} \times \frac{\sin{x}}{x}\times \frac{x}{1+\cos x} = 0 . \end{eqnarray*} $$

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\begin{align*} L = \lim_{x \to 0} \frac{1-\cos{x}}{x} &= \lim_{x \to 0} \frac{\sin^{2}{x} + \cos^{2}{x} -\cos{x}}{x} \\ &= \lim_{x \to 0} \cos(x) \cdot \frac{\cos{x}-1}{x} + \lim_{x \to 0} \frac{\sin^{2}{x}}{x} \\ &=-L \Rightarrow 2L =0 \Rightarrow L=0 \end{align*}

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Aren't you presupposing the limit exists? – David Mitra Nov 13 '11 at 21:48

Using L'Hopital

$$\lim \limits_{x \to 0} \frac{1 - \cos x}{x} = \lim \limits_{x \to 0}\frac{(1 - \cos x)'}{x'} = \lim \limits_{x \to 0} \sin x = 0$$

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