Evaluating the limand directly at x = 0 yields the indeterminate form 0/0. If L'Hopital's rule is known to you, you can compute the limit by applying it twice:
[tex]\displaystyle\lim_{x\to0}\frac{x\left(e^{3x}-1\right)}{2-2\cos(x)} = \lim_{x\to0}\frac{3xe^{3x}+e^{3x}-1}{2\sin(x)} \\\\\\ = \lim_{x\to0}\frac{9xe^{3x}+6e^{3x}}{2\cos(x)} = \frac62=\boxed{3}[/tex]
