# How do you evaluate the limit (3(1-cosx))/x as x approaches 0?

Sep 2, 2016

$0$

#### Explanation:

${\lim}_{x \to 0} \frac{3 \left(1 - \cos x\right)}{x}$

we're going to steer this in the direction of well known limit ${\lim}_{z \to 0} \frac{\sin z}{z} = 1$

using the double angle identity: $\cos 2 A = 1 - {\sin}^{2} A$

$= {\lim}_{x \to 0} \frac{3 \left(1 - \left(1 - 2 {\sin}^{2} \left(\frac{x}{2}\right)\right)\right)}{x}$

$= {\lim}_{x \to 0} \frac{6 {\sin}^{2} \left(\frac{x}{2}\right)}{x}$

pattern matching numerator and denominator, and taking the constant term outside the limit
$= 12 {\lim}_{x \to 0} \frac{{\sin}^{2} \left(\frac{x}{2}\right)}{\frac{x}{2}}$

$= 12 {\lim}_{x \to 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \sin \left(\frac{x}{2}\right)$

these limits of these individual terms exist and the limit of the product is the product of the limits

$= 12 {\lim}_{x \to 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot {\lim}_{x \to 0} \sin \left(\frac{x}{2}\right)$

$= 12 \cdot 1 \cdot 0 = 0$

You cannot use L'Hopital's Rule for this.