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 that
\begin{align}
\lim_{x \rightarrow 0} \frac{\sin 2x}{x^3} + a + \frac{b}{x^2} 
& = \lim_{x \rightarrow 0} \ a + \frac{\sin 2x + bx}{x^3} \\
& = \lim_{x \rightarrow 0} \ a + \frac{2 \cos 2x + b}{3x^2},
\end{align}
which gives that $b=-2$, otherwise the second term blows up. Continuing,
\begin{align}
\lim_{x \rightarrow 0} \ a + \frac{2 \cos 2x -2 }{3x^2}
& = \lim_{x \rightarrow 0} \ a + \frac{-4 \sin 2x}{6x} \\
& = \lim_{x \rightarrow 0} \ a + \frac{-8 \cos 2x}{6} \\
& = \lim_{x \rightarrow 0} \ a - \frac{8}{6}.
\end{align}
Hence $a = 8/6$ and $3a+b = 2$.  
