For what values of a and b is the following equation true? lim x →...

We have
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{\sin (2x)}{x^3}+a+\frac{b}{x^2}\\ & =\underset{x\to 0}{lim}\frac{\sin (2x)+a{x}^3+bx}{x^3}\end{array}$
which is a limit of type $0/0$, so we try L'Hôpital's rule:
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{2\cos (2x)+3a{x}^2+b}{3x^2}\end{array}$
The denominator again vanishes for $x\to 0$ , the nominator goes to $2+b$. So if $b\ne -2$, the nominator does not vanish and we have
$L=\mathrm{sgn}(2+b)\mathrm{\infty }$
For $b=-2$ we again have a limit of type $0/0$ and apply the rule again:
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{-4\sin (2x)+6ax}{6x}\end{array}$
The nominator and denominator vanish and we apply the rule once again:
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{-8\cos (2x)+6a}{6}=\frac{6a-8}{6}=a-\frac{4}{3}\end{array}$
This gives the answer that the equation is true, L vanishes, if $a=4/3$ and $b=-2$