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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...

  • 0
    More solutions to this limit are [here](http://math.stackexchange.com/q/420698/43351) and [here](http://math.stackexchange.com/q/36299/43351).2013-06-15

4 Answers 4

12

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*} $

2

\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 e$x$ists?2011-11-13
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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