$\displaystyle \lim_{x\to 0}\frac{(sin 2x)}{(x^3)}+a+(\frac{b}{x^3})=0$

The problem states "find a and b such that the limit as x approaches 0 is equal to 0."

My strategy has been to use L'Hospital's Rule to evaluate the first term, ((sin 2x)/x^3), and then find an a and b that cancel out the leading term near x=0.

OK so my first problem is evaluating the leading term as x approaches 0. Since sine is positive between 0 and 2 and x cubed is positive , for negative x the quotient is positive as well, then the limit of the first term as x approaches 0 has to be positive infinity. I am at a loss as to how I should go about showing that this is true. (applying L'Hospital's leads to 2/0 so that doesn't seem to be too fruitful. I have tried substituting [2 cos x sin x])

The second difficulty is finding an a and b that cancel out the leading term. I think I need a and b such that their limit as x approaches 0 is not only negative infinity, but they have to decrease at the same rate that the leading term increases.

Perhaps?

$\displaystyle \frac{-(sin 2x)}{(x^3)}= \frac{-1*(sin 2x)}{x^3} + \frac{-1}{x^3}$

$\displaystyle a = \frac{-1*(sin 2x)}{x^3}$

b = -1

This seems really stupid though...

I'll be trying to figure out latex over the next 15min or so. Just looking to clean up the problem statement.