# Question #a5dc1

Nov 24, 2016

#### Explanation:

Prove: ${\lim}_{x \to 0} \frac{\sin \left(6 x\right) + \sin \left(8 x\right)}{\sin \left(4 x\right) + \sin \left(6 x\right)} = \frac{7}{5}$

The expression evaluated at the limit is the indeterminate form $\frac{0}{0}$, therefore, one should use L'Hôpital's rule

Take the derivative of the numerator:

$\frac{d \left\{\sin \left(6 x\right) + \sin \left(8 x\right)\right\}}{\mathrm{dx}} = 6 \cos \left(6 x\right) + 8 \cos \left(8 x\right)$

Take the derivative of the denominator:

$\frac{d \left\{\sin \left(4 x\right) + \sin \left(6 x\right)\right\}}{\mathrm{dx}} = 4 \cos \left(4 x\right) + 6 \cos \left(6 x\right)$

Make a new fraction with the new numerator and denominator:

${\lim}_{x \to 0} \frac{6 \cos \left(6 x\right) + 8 \cos \left(8 x\right)}{4 \cos \left(4 x\right) + 6 \cos \left(6 x\right)}$

The above can be evaluated at the limit:

$\frac{6 \cos \left(6 \left(0\right)\right) + 8 \cos \left(8 \left(0\right)\right)}{4 \cos \left(4 \left(0\right)\right) + 6 \cos \left(6 \left(0\right)\right)} = \frac{14}{10} = \frac{7}{5}$

The rules says that this is the limit of the original expression:

${\lim}_{x \to 0} \frac{\sin \left(6 x\right) + \sin \left(8 x\right)}{\sin \left(4 x\right) + \sin \left(6 x\right)} = \frac{7}{5}$

Q.E.D.