Finding $\lim_{x\to0}\frac{1-\cos(x)}{x}$ with squeeze theorem

Question

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.

Answer 1

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 $\displaystyle AB$.

Answer 2

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 $0$.

Answer 3

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

Answer 4

Using that $|\sin x| \le |x|$

$$\frac{1-\cos x}{|x|}=\frac{\sin^2 x}{|x|(1+\cos x)} \le \frac{|x|}{1+\cos x}\to 0$$