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.
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2$\begingroup$ 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. $\endgroup$ – Soarer May 1 '11 at 23:27
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$\begingroup$ @Soarer I want to the squeeze theorem because its the requirement of the homework. $\endgroup$ – backus May 1 '11 at 23:40
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$\begingroup$ More solutions to this limit are here and here. $\endgroup$ – Lord_Farin Jun 15 '13 at 7:56
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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 $\displaystyle AB$.
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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 $0$.
answered May 1 '11 at 23:15
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$\begingroup$ 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. $\endgroup$ – Soarer May 1 '11 at 23:29
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$\begingroup$ Is there a way you could solve this without expanding cos(x) to that series? $\endgroup$ – backus May 2 '11 at 1:00
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For any $x$ such that $0 < |x| \leq \pi/2$, $$ 0 \le \bigg|\frac{{1 - \cos x}}{x}\bigg| = \frac{{1 - \cos |x|}}{{|x|}} = \frac{{\int_0^{|x|} {\sin t \,dt} }}{{|x|}} \le \frac{{\int_0^{|x|} {\sin |x| \, dt} }}{{|x|}} = \sin |x|. $$
