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.)

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:
$$ \lim\limits_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}= 0 , a = \frac{4}{3}, b = -2$$
