first post

Hello and welcome, Mike!

I recommend in the future posting your problems in the appropriate forums, as they will receive more and prompt attention that way.

I think your first move is a good one, i.e., to write:

$\lim_{x\to0^{+}}\frac{\sin(x)+ax^3+bx}{x^3}=0$

We have the indeterminate form 0/0, so application of L'Hôpital's rule gives:

$\lim_{x\to0^{+}}\frac{\cos(x)+3ax^2+b}{3x^2}=0$

In order for this limit to be bounded, we require the numerator to tend to 0, so we need $b=-1$ , and we may write:

$\lim_{x\to0^{+}}\frac{\cos(x)+3ax^2-1}{3x^2}=0$

We have again the indeterminate form 0/0, so application of L'Hôpital's rule gives:

$\lim_{x\to0^{+}}\frac{-\sin(x)+6ax}{6x}=0$

We have again the indeterminate form 0/0, so application of L'Hôpital's rule gives:

$\lim_{x\to0^{+}}\frac{-\cos(x)+6a}{6}=0$

Can you now finish?