L'Hopital's Rule Values for an equation I have the following task:

For what values of $a$ and $b$ is the following equation true?
  $$\lim\limits_{x \to 0} \left(\frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}\right) = 0 $$

I want to know the steps I should follow in order to find the solution.

We have
\begin{align}
L 
&= \lim_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2} \\
&= \lim_{x \to 0} \frac{\sin(2x) + a x^3 + b x }{x^3}\\
\end{align} 
which is a limit of type $0 / 0$, so we try L'Hôpital's rule:
\begin{align}
L &= \lim_{x \to 0} \frac{2\cos(2x) + 3 a x^2 + b}{3x^2}
\end{align}
The denominator again vanishes for $x \to 0$, the nominator goes to
$2 + b$.
So if $b \ne -2$, the nominator does not vanish and we have 
$$
\DeclareMathOperator{sgn}{sgn}
L = \sgn(2 + b) \, \infty
$$
For $b = -2$ we again have a limit of type $0 / 0$ and apply the rule again:
\begin{align}
L 
&= \lim_{x\to 0} \frac{-4 \sin(2x) + 6ax}{6x}
\end{align}
The nominator and denominator vanish and we apply the rule once again:
\begin{align}
L 
&= \lim_{x\to 0} \frac{-8 \cos(2x) + 6a}{6} = \frac{6a-8}{6}
= a - \frac{4}{3}
\end{align}
This gives the answer that the equation is true, $L$ vanishes, 
if $a = 4/3$ and $b = -2$.
