# Finding the limit of $(1-\cos(x))/x$ as $x\to 0$ with squeeze theorem

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.

• Is there any particular reason on using squeeze theorem? Multiplying $1+ \cos (x)$ in both numerator and denominator would be something more natural to do to me. – Soarer May 1 '11 at 23:27
• @Soarer I want to the squeeze theorem because its the requirement of the homework. – backus May 1 '11 at 23:40
• More solutions to this limit are here and here. – Lord_Farin Jun 15 '13 at 7:56

Here is a geometric squeeze: 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 $\displaystyle AB$.

• Did you draw this image? If so, how did you do it? – JavaMan May 2 '11 at 2:51
• @DJC: Yes, using mspaint... – Aryabhata May 2 '11 at 2:54
• @DJC: I have updated the figure using one drawn with GeoGebra :-) – Aryabhata Jul 15 '11 at 5:54
• Cool. Thanks! (more letters) – JavaMan Jul 15 '11 at 12:50

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 $0$.

• The correcct upper bound is $1 - cos (x) < 1 - 1 + x^2/2$ for $x$ close to 0. This can be seen by multiplying $\frac{1+\cos (x)}{1+\cos (x)}$, and bounding $1+\cos (x)$ away from 0 for $x$ close to 0. – Soarer May 1 '11 at 23:29
• @Soarer: Yes, I got that already. – Alex Becker May 1 '11 at 23:32
• Is there a way you could solve this without expanding cos(x) to that series? – backus May 2 '11 at 1:00

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