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

Solution 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 $\displaystyle AB$.

Solution 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 $0$.