Tricky Limits Problem () Problem:
If $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=0$ then find the value of $3a+b$.
My attempt: $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=\lim_{x \to 0}{\sin2x\over 2x}({2\over x^2})+a+{b\over x^2}={2+b+ax^2\over x^2}$.From this we can conclude that $a=0$ and $b=-2$, hence $3a+b=-2$. However the answer is $2$. Where am I going wrong?

we have
$sin(X)=X-\frac{X^3}{6}+X^3\epsilon(X)$
with $\displaystyle{  \lim_{X\to 0}  \epsilon(X)=0}$.
thus
$sin(2x)=2x-\frac{4x^3}{3}+x^3\epsilon(x)$.
so
$\frac{sin(2x)}{x^3}+a+\frac{b}{x^2}$
$=\frac{2}{x^2}-\frac{4}{3}+a+\frac{b}{x^2}+\epsilon(x)$.
the limit is $0$ if
$a=\frac{4}{3}$ and $b=-2$.
finally
we will have
$3a+b=2$.
