Another Limit Conundrum 
For what values of $a$ and $b$ is the following limit true?
  $$\lim_{x\to0}\left(\frac{\tan2x}{x^3}+\frac{a}{x^2}+\frac{\sin bx}{x}\right)=0$$

This question is really confusing me. I know that $\tan(2x)/x^3$ approaches infinity as $x$ goes to $0$ (L'Hôpital's). I also understand that $\sin(bx)/x$ goes to $b$ as $x$ approaches $0$. However, I am not sure how to get rid of this infinity with the middle term. Any ideas?
Thanks!
 A: Since $\lim_{x\to 0}\dfrac{\sin bx} {x} =b$ the given condition is equivalent to $$\lim_{x\to 0}\frac{\tan 2x+ax}{x^3}=-b$$ or $$\lim_{x\to 0}\frac{\tan 2x-2x}{x^3}+\frac{a+2}{x^2}=-b$$ Substituting $t=2x$ we get $$\lim_{t\to 0}8\cdot \frac{\tan t-t} {t^3}+\dfrac{4a+8}{t^2}=-b$$ The first fraction tends to $1/3$ (as can be easily seen via L'Hospital's Rule or Taylor series) and therefore the above equation is equivalent to $$\lim_{t\to 0}\frac{4a+8}{t^2}=-b-\frac{8}{3}$$ Multiplication by $t^2$ now gives us $4a+8=0$ or $a=-2$ and then from the above equation we get $b=-8/3$.
A: First, notice that your limit exists if and only if it exists without the last term (since, as you said, the limit of the last term exists). 
You have 
$$\tag1
\frac{\tan2x}{x^3}+\frac a{x^2}=\frac{\tan2x+ax}{x^3}.
$$
Note that if $a\ne0$ you cannot apply L'Hôpital. The expression in $(1)$ is, close to $0$, 
$$\tag2
\frac{\tan2x+ax}{x^3}=\frac{x+o(x^3)+ax}{x^3}.
$$
This requires $a=-1$ for the limit to exist. In more detail, 
$$\tag2
\frac{\tan2x+ax}{x^3}=\frac{2x+\tfrac{(2x)^3}3+o(x^5)+ax}{x^3}=\frac{2+a}{x^2}+\tfrac83+o(x^2).
$$
So it converges to $\tfrac13$ when $a=-2$. For the whole limit to be zero, we need that $$\tag3\lim_{x\to0}\frac{\sin bx}{x}=-\frac83.$$
In the end, we need $a=-2$, $b=-\tfrac83$. 
A: Using the usual Taylor series for
$$y=\frac{\tan(2x)}{x^3}+\frac{a}{x^2}+\frac{\sin (bx)}{x}=\frac{\tan(2x)+ax+x^2\sin (bx)}{x^3}$$ The numerator write
$$\left(2 x+\frac{8 x^3}{3}+\frac{64 x^5}{15}+O\left(x^7\right) \right)+a x+x^2\left(b x-\frac{b^3 x^3}{6}+\frac{b^5 x^5}{120}+O\left(x^7\right) \right)$$ that is to say
$$(a+2) x+\left(b+\frac{8}{3}\right) x^3+\left(\frac{64}{15}-\frac{b^3}{6}\right)
   x^5+O\left(x^7\right)$$ making $$y=\frac{a+2}{x^2}+\left(b+\frac{8}{3}\right)+\left(\frac{64}{15}-\frac{b^3}{6}\right)
   x^2+O\left(x^4\right)$$ So, in order to have a limit equal to $0$, you need $a=-2$, $b=-\frac{8}{3}$ and then
$$y=\frac{3008 }{405}x^2+O\left(x^4\right)$$
A: We have that
$$\frac{\tan (2x)}{x^3}+\frac{a}{x^2}+\frac{\sin bx}{x}=\frac{\tan (2x)+ax+x^2\sin bx}{x^3}$$
and from here since $x^2\sin bx\sim bx^3$ and $\tan (2x)\sim 2x$ we need $a=-2$ as a necessary condition for the limit to exist then we can consider by standard limits
$$\frac{\tan (2x)-2x+x^2\sin bx}{x^3}=8\frac{\tan (2x)-2x}{(2x)^3}+b\frac{\sin bx}{bx}\to \frac83+b=0$$
