# How do you find the limit of  (1-cos(x))/x as x approaches 0?

Feb 22, 2016

0

#### Explanation:

When in doubt, use L'Hopitals rule, maybe even more then once, especially if there are polynomials on the bottom or constants to be rid of, make them all go poof!

L'Hopitals rule says that:
${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

We trying to solve,
${\lim}_{x \to 0} \frac{1 - \cos \left(x\right)}{x}$.

Take the derivative of the top and the bottom functions:
$\frac{d}{\mathrm{dx}} \left(1 - \cos \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(1\right) - \frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = 0 - \left(- \sin \left(x\right)\right) = \sin \left(x\right)$
and $\frac{d}{\mathrm{dx}} \left(x\right) = 1$.
So, ${\lim}_{x \to 0} \frac{1 - \cos \left(x\right)}{x} = {\lim}_{x \to 0} \frac{\sin \left(x\right)}{1} = {\lim}_{x \to 0} \sin \left(x\right)$
Since $\sin \left(x\right)$ is well behaved at zero, this is just
$\sin \left(0\right)$ which is $0$.