# What is the limit of (cos(3x)-cos(4x))/x^2 as x approaches 0?

May 7, 2018

${\lim}_{x \rightarrow 0} \frac{16 \cos \left(4 x\right) - 9 c o \left(3 x\right)}{2} = \frac{16 - 9}{2} = \frac{7}{2}$

#### Explanation:

show below

${\lim}_{x \rightarrow 0} \frac{\cos \left(3 x\right) - \cos \left(4 x\right)}{x} ^ 2$

Direct compensation product equal $\frac{0}{0}$ so we will use l'hospital rule

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = \frac{0}{0}$

${\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

$f \left(x\right) = \cos \left(3 x\right) - \cos \left(4 x\right)$

$f ' \left(x\right) = 4 \sin \left(4 x\right) - 3 \sin \left(3 x\right)$

$g \left(x\right) = {x}^{2}$

$g ' \left(x\right) = 2 x$

${\lim}_{x \rightarrow 0} \frac{4 \sin \left(4 x\right) - 3 \sin \left(3 x\right)}{2 x}$

since the Direct compensation product again equal $\frac{0}{0}$ so we will derive it again

$f ' ' \left(x\right) = 16 \cos \left(4 x\right) - 9 c o \left(3 x\right)$

$g ' ' \left(x\right) = 2$

${\lim}_{x \rightarrow 0} \frac{16 \cos \left(4 x\right) - 9 c o \left(3 x\right)}{2} = \frac{16 - 9}{2} = \frac{7}{2}$