# What is the limit as x approaches 0 of sin(3x)/sin(4x)?

The answer is $\frac{3}{4}$.
You both $\sin \left(3 \cdot 0\right) = 0$ and $\sin \left(4 \cdot 0\right) = 0$, so you can use l'hopitals rule. This is:
${\lim}_{x \to 0} \sin \frac{3 x}{\sin} \left(4 x\right) = {\lim}_{x \to 0} \frac{\left[\sin \left(3 x\right)\right] '}{\left[\sin \left(4 x\right)\right] '} = {\lim}_{x \to 0} \frac{3 \cos \left(3 x\right)}{4 \cos \left(4 x\right)} = \frac{3 \cos \left(0\right)}{4 \cos \left(0\right)} = \frac{3}{4}$
The first step is taking the derivative of both the nominator and the denominator; the last step is just filling in zero, you're allowed to do this because $\cos \left(0\right) = 1$, so you don't risk dividing by zero.