<s>[INST] 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!
 [/INST] 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
$$y=\frac{3008 }{405}x^2+O\left(x^4\right)$$
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