For what value(s) of $a,b$ does this limit hold? 
For which value(s) of $a$ and $b$ is $$ \lim_{x \to 0}(x^{-3}\sin(3x)+ax^{-2}+b)=0$$  

My Attempt
Firstly I rewrote it into this form;
$$ \lim_{x \to 0}\left(\dfrac{\sin(3x)}{x^3}+\dfrac{a}{x^{2}}+b\right)=0$$
Then by putting them all under one common denominator,
$$ \lim_{x \to 0}\left(\dfrac{\sin(3x)+xa+x^2b}{x^3}\right)=0$$
So the above is the identity we want to hold; 
If we just look at the limit;
$$ \lim_{x \to 0}\left(\dfrac{\sin(3x)+xa+x^2b}{x^3}\right)$$ we can see L'Hopital can be used on this as for all  $a,b \in \mathbb{R}$ this is a $0/0$ type limit; differentiating gives;
$$\left(\dfrac{3\cos(3x)+a+2xb}{3x^2}\right)$$
Now I am stuck on how to decide which value of $a,b$ hold; if $a=-3$ and for all values of $b$ at $x=0$ the above fraction would be of the form $0/0$ would I then differentiate again?  
Any help would be appreciated
 A: 
Then by putting them all under one common denominator,
  $$ \lim_{x \to 0}\left(\dfrac{\sin(3x)+xa+x^2b}{x^3}\right)=0$$   

You have made a mistake here; $x^2b$ should be $x^3b$.
For the next part, you correctly use L'Hopital, but with the $x^3b$ term it gives you the slightly different answer
$$
\lim_{x \to 0}\left(\dfrac{3\cos(3x)+a+3x^2b}{3x^2}\right)
$$

Now I am stuck on how to decide which value of $a,b$ hold; if $a=-3$ and for all values of $b$ at $x=0$ the above fraction would be of the form $0/0$ would I then differentiate again?  

Yes, exactly.  Since the denominator approaches $0$, you need the numerator to also approach $0$ if the limit is to exist.  So $a = -3$, and then you differentiate again to get
$$
\lim_{x \to 0}\left(\dfrac{-9\sin(3x) + 6xb}{6x}\right)
$$
Apply L'Hopital again...
$$
\lim_{x \to 0}\left(\dfrac{-27\cos(3x) + 6b}{6}\right)
$$
You don't need to apply L'Hopital anymore, as the denominator is now nonzero.
But you need this limit to evaluate to $0$.  So what is $b$?
A: For the above limit to exist, $a$ must be equal to $-3$.  Putting $a=-3$, you again have a $\frac{0}{0}$ form and hence you apply L'hopital to reach your answer.
Hope this was helpful. :)
