For what values of a and b does the following limit equal 0? 
I understand I need to make the sum of the individual limits equal 0 - but I'm a little lost. I computed the limit of the first term to be -4/3 via L'Hospitals Rule but Wolframalpha contradicts me (http://www.wolframalpha.com/input/?i=limit+x-%3E+0+%28sin%282x%29%2F%28x%5E3%29%29). 
'a' obviously should be left for last - and $b/($x^2) is some constant - so I get 0? Assuming 'b' is positive, or anything other than 0, I get that term is 0.
Thus, -4/3 + 0 + a = 0
Simple algebraic manipulating would lead me to believe a = 4/3 and then I'd plug that back in to find 'b'. 
Questions
Why does Wolfram state what it does?
Is this solution correct? (I don't have the answer for the problem.)
 A: We have by the Taylor series
$$\sin(2x)=2x-\frac{4}{3}x^3+o(x^3)$$
so it's clear that for $b=-2$ and $a=\frac{4}{3}$ the limit is $0$.
A: First let us put this into a better form, with one variable term and one constant term:
$$ \lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} + a = 0$$ 
well if we evaluate the limit using L'Hopitals we get:
$$\lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} = \lim\limits_{x \to 0} \frac{2\cos(2x) + b}{3x^2} $$
$$\lim\limits_{x \to 0} \frac{2\cos(2x) + b}{3x^2} = \frac{2+b}{0} = (2+b) \infty $$
Since this technically equals infinity in all cases except when $(2+b) = 0$, we make the $b = -2$ so that our limit comes out to be zero. Now let's figure out the $a$
First I want to plug in the $b$ back into our original equation:
$$ \lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} + a =  \lim\limits_{x \to 0} \frac{\sin(2x) - 2x}{x^3} + a = 0$$
then let's take the limit
$$ \lim\limits_{x \to 0} \frac{\sin(2x) - 2x}{x^3} + a = \lim\limits_{x \to 0} \frac{2\cos(2x) - 2}{3x^2} + a = \frac{2}{3}\lim\limits_{x \to 0} \frac{\cos(2x) - 1}{x^2} + a$$
and we can apply the L'Hospitals rule again:
$$ \frac{2}{3}\lim\limits_{x \to 0} \frac{\cos(2x) - 1}{x^2} + a = \frac{2}{3} \lim\limits_{x \to 0} \frac{-2\sin(2x)}{2x} + a =  \frac{2}{3} \lim\limits_{x \to 0} \frac{-\sin(2x)}{x} + a$$
and once again with L'Hospitals:
$$ \frac{2}{3} \lim\limits_{x \to 0} \frac{-\sin(2x)}{x} +a = \frac{2}{3} \lim\limits_{x \to 0} {-2\cos(2x)} + a = \frac{-4}{3} + a$$
If we then set this equal to zero we get $a = \frac{4}{3} $
And so our final answer is:
$$ \lim\limits_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}= 0 , a = \frac{4}{3}, b = -2$$
A: Hint: you're correct that $a$ can be taken care of when you have found
$$
\lim_{x\to0}\frac{\sin(2x)+bx}{x^3}
$$
that you can compute by applying l'Hôpital's theorem. But you have to compute derivatives right.

For instance, just to show your computations were wrong,
$$
\lim_{x\to0}\frac{\sin(2x)}{x^3}=
\lim_{x\to0}\frac{2\cos(2x)}{3x^2}=\infty
$$
You can't go on with l'Hôpital here, because it's not a $0/0$ form any more.
