# How do you find the limit of (1-cos(4x))/(1-cos(3x) as x approaches 0?

May 17, 2016

${\lim}_{x \to 0} \frac{1 - \cos \left(4 x\right)}{1 - \cos \left(3 x\right)} = \frac{16}{9}$

#### Explanation:

The series expansion of $\cos \left(\alpha x\right)$ for $x = 0$ in an open set containing $0$ is given bysum_{i=0}^{infty} (-1)^i((alpha x)^{2i})/(2n!)

Then
$1 - \cos \left(4 x\right) = 8 {x}^{2} - \frac{32 {x}^{4}}{3} + \frac{256 {x}^{6}}{45} + \ldots$
also
$1 - \cos \left(3 x\right) = \frac{9 {x}^{2}}{2} - \frac{27 {x}^{4}}{8} + \frac{81 {x}^{6}}{80} + \ldots$
so
$1 - \cos \left(4 x\right) = {x}^{2} \left(8 - \frac{32 {x}^{2}}{3} + \frac{256 {x}^{4}}{45} + \ldots\right)$
also
$1 - \cos \left(3 x\right) = {x}^{2} \left(\frac{9}{2} - \frac{27 {x}^{2}}{8} + \frac{81 {x}^{4}}{80} + \ldots\right)$

Substituting in the fraction
$\frac{1 - \cos \left(4 x\right)}{1 - \cos \left(3 x\right)} = \frac{\left(8 - \frac{32 {x}^{2}}{3} + \frac{256 {x}^{4}}{45} + \ldots\right)}{\left(\frac{9}{2} - \frac{27 {x}^{2}}{8} + \frac{81 {x}^{4}}{80} + \ldots\right)}$
computing the limit for $x \to 0$ we obtain
${\lim}_{x \to 0} \frac{1 - \cos \left(4 x\right)}{1 - \cos \left(3 x\right)} = \frac{16}{9}$