# Limit as x approaches 0 of sin(4x)/sin(6x)?

Dec 14, 2014

This limit gives a 0/0 indeterminate form but you can use de l'Hospital Rule to get the result of 4/6.
De l'Hospital Rule is used to solve this kind of problems by deriving the nominator and denominator of the fraction and then do the limit of the new function obtained.
Deriving you get:

$f \left(x\right) = \frac{4 \cos \left(4 x\right)}{6 \cos \left(6 x\right)}$ when x->0 you get 4/6.

Dec 14, 2014

It's $\frac{4}{6}$.

Evaluating this limit by substitution gives us the indeterminate form $\frac{0}{0}$.
However, we can use de l'Hospital Rule, by differentiating the numerator and denominator of the fraction and then evaluating the limit of the new fraction obtained, as follows:

Differentiating the numerator and the denominator, via the chain rule:
${\lim}_{x \to 0} \sin \frac{4 x}{\sin} \left(6 x\right) = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left[\sin \left(4 x\right)\right]}{\frac{d}{\mathrm{dx}} \left[\sin \left(6 x\right)\right]} = {\lim}_{x \to 0} \frac{4 \cos \left(4 x\right)}{6 \cos \left(6 x\right)}$

Evaluating this limit by substitution:

${\lim}_{x \to 0} \sin \frac{4 x}{\sin} \left(6 x\right) = \frac{4 \cos \left(4 \times 0\right)}{6 \cos \left(6 \times 0\right)} = \frac{4 \times 1}{6 \times 1} = \frac{4}{6}$