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How do I find:

$ \lim_{x\to0}\frac{1-\cos(x)}{x} $

Using the squeeze theorem. Particularly, how would I arrive at its bounding functions?

If possible, please try not to use derivatives.

  • 0
    More solutions to this limit are [here](http://math.stackexchange.com/q/81768/43351) and [here](http://math.stackexchange.com/q/420698/43351).2013-06-15

3 Answers 3

16

Here is a geometric squeeze:

enter image description here

Now we can show that:

$ \frac{1 - \cos x}{x} \lt \sin \frac{x}{2}$

We have $\displaystyle y = x/2$ ($\triangle ABC$ is isosceles and so $\angle CAB = \frac{\pi - x}{2}$) and hence in $\triangle BDA$, $\displaystyle \sin \frac{x}{2} = \frac{AD}{AB} \gt \frac{AD}{x}$, as $\displaystyle x$ is the length of the arc $\displaystyle AB$.

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    Cool. Thanks! (more letters)2011-07-15
5

We can write $\cos(x)$ as $1 - x^2/2 + x^4/6 - \cdots$ and so near $x = 0$ we have $1 - \cos(x) < x^2/2$, and so $\frac{1 - \cos(x)}{x} < \frac{x^2}{2x} = \frac{x}{2}$. At the same time, $\cos(x) < 1$ so $\frac{1 - \cos(x)}{x} > 0$. Thus we have $\lim_{x\to0}0 \leq \lim_{x\to0}\frac{1-\cos(x)}{x} \leq \lim_{x\to0}\frac{x}{2}$, which by the squeeze theorem is $0$.

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    Is there a way you could solve this without expanding cos(x) to that series?2011-05-02
2

For any $x$ such that $0 < |x| \leq \pi/2$, $ 0 \le \bigg|\frac{{1 - \cos x}}{x}\bigg| = \frac{{1 - \cos |x|}}{{|x|}} = \frac{{\int_0^{|x|} {\sin t \,dt} }}{{|x|}} \le \frac{{\int_0^{|x|} {\sin |x| \, dt} }}{{|x|}} = \sin |x|. $