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 $3a+b=2$.
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 $3a+b = 2$.
