Question

Without the L'hopitals stuff we have not gone over that yet.

Answer 1

You may have already learned that

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax} = 1$

for a ≠ 0.

Then for your limit, you can write

$\displaystyle \lim_{x\to0}\frac{\sin(2x)}{3x} = \frac23 \lim_{x\to0} \frac{\sin(2x)}{2x} = \frac23\cdot1 = \boxed{\frac23}$