How do you evaluate the limit (1-cosx)/x as x approaches 0?

Sep 26, 2016

Because the expression evaluated at the limit is $\frac{0}{0}$, we can use L'Hospital's Rule.

Explanation:

The rule states that, when given the limit of a fraction, $\frac{\lim}{x \rightarrow c} f \frac{x}{g} \left(x\right)$ where $f \frac{c}{g} \left(c\right)$ is, $\frac{0}{0}$, $\frac{\infty}{\infty}$ or $\frac{- \infty}{-} \infty$, then $\frac{\lim}{x \rightarrow c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ goes to the same limit.

To implement the rule, we take the derivative of the numerator:
$d \frac{1 - \cos \left(x\right)}{\mathrm{dx}} = \sin \left(x\right)$

the derivative of the of the denominator:

$\frac{\mathrm{dx}}{\mathrm{dx}} = 1$

Assemble the new fraction with the same limit:

$\frac{\lim}{x \rightarrow 0} \sin \frac{x}{1} = 0$

Therefore, $\frac{\lim}{x \rightarrow 0} \frac{1 - \cos \left(x\right)}{x} = 0$

Sep 26, 2016

Usually, this is done after showing that ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$. See below.

Explanation:

$\frac{1 - \cos x}{x} = \frac{\left(1 - \cos x\right)}{x} \cdot \frac{\left(1 + \cos x\right)}{\left(1 + \cos x\right)}$

$= \frac{1 - {\cos}^{2} x}{x \left(1 + \cos x\right)}$

$= {\sin}^{2} \frac{x}{x \left(1 + \cos x\right)}$

$= \sin \frac{x}{x} \cdot \sin x \cdot \frac{1}{1 + \cos x}$

Taking the limit as $x \rightarrow 0$, we get

$\left(1\right) \left(0\right) \left(\frac{1}{2}\right) = 0$