| $\displaystyle \lim_{x \to 0} \frac{1-\cos x}{x}$ |
$=\displaystyle \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x)}{x(1+\cos x)}$ |
| |
|
| |
$=\displaystyle \lim_{x \to 0} \frac{1-\cos x+\cos x-\cos^2 x}{x(1+\cos x)}$ |
| |
|
| |
$=\displaystyle \lim_{x \to 0} \frac{1-\cos^2 x}{x(1+\cos x)}$ |
| |
|
| |
$=\displaystyle \lim_{x \to 0} \frac{\sin^2 x}{x(1+\cos x)}$ |
| |
|
| |
$=\displaystyle \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{\sin x}{1+\cos x}$ |
| |
|
| |
$=\displaystyle \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \left( \lim_{x \to 0} \frac{\sin x}{1+\cos x} \right)$ by this entry |
| |
|
| |
$=\displaystyle 1 \cdot \lim_{x \to 0} \frac{\sin x}{1+\cos x}$ by this entry |
| |
|
| |
$=\displaystyle \frac{\sin 0}{1+\cos 0}$ by this entry and the fact that $\sin$ and $\cos$ are continuous |
| |
|
| |
$=\displaystyle \frac{0}{1+1}$ |
| |
|
| |
$=0$ |
![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
