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.
 A: 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 $b = -2$.
A: If you use L'Hôpital's rule on $\frac{\sin(2x)}{x^3}$ and $\frac{b}{x^2}$ you will see that neither has a limit as $x$ approaches zero. Therefore you cannot split your full limit into separate limits to get your answer.
Put the expression inside the limit into a single fraction, with denominator $x^3$. Ensure that the expression has the form $\frac 00$. Then use L'Hôpital's rule to find its limit. If that does not have a clear limit, again make sure the expression has a proper form for L'Hôpital's rule, which may put limits on $a$ and/or $b$, then use L'Hôpital's rule again.
Keep doing this until you get a clear limit involving $a$ and/or $b$. Set that limit to zero and solve for $a$ and/or $b$.
A: $\lim_{x\to0}\frac{\sin 2x}{2x}=1$.  The rest of the terms   $\frac{2}{x*x}+a+\frac{b}{x*x}=\frac{2+b}{x*x}+a=0$
So $b=-2$, $a=0$
