Answer
$\lim\limits_{x \to 0} \frac{sin(3x)}{tan(3x)} =1$
Work Step by Step
Using the rules of limits, we find:
$$\lim\limits_{x \to 0} \frac{sin(3x)}{tan(3x)}\\ = \lim\limits_{x \to 0} \frac{sin(3x)cos(3x)}{sin(3x)} \\= \lim\limits_{x \to 0} \frac{cos(3x)}{1} \\= cos(3*0) = 1$$
