Krdytkyu

For what values of $a$ and $b$ is the following limit true?
$$lim_{xto0}left(frac{tan2x}{x^3}+frac{a}{x^2}+frac{sin bx}{x}right)=0$$

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!

Since $lim_{xto 0}dfrac{sin bx} {x} =b$ the given condition is equivalent to $$lim_{xto 0}frac{tan 2x+ax}{x^3}=-b$$ or $$lim_{xto 0}frac{tan 2x-2x}{x^3}+frac{a+2}{x^2}=-b$$ Substituting $t=2x$ we get $$lim_{tto 0}8cdot frac{tan t-t} {t^3}+dfrac{4a+8}{t^2}=-b$$ 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 $$lim_{tto 0}frac{4a+8}{t^2}=-b-frac{8}{3}$$ Multiplication by $t^2$ now gives us $4a+8=0$ or $a=-2$ and then from the above equation we get $b=-8/3$.

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 $ane0$ 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 $$tag3lim_{xto0}frac{sin bx}{x}=-frac83.$$
In the end, we need $a=-2$, $b=-tfrac83$.

Using the usual Taylor series for
$$y=frac{tan(2x)}{x^3}+frac{a}{x^2}+frac{sin (bx)}{x}=frac{tan(2x)+ax+x^2sin (bx)}{x^3}$$ The numerator write
$$left(2 x+frac{8 x^3}{3}+frac{64 x^5}{15}+Oleft(x^7right) right)+a x+x^2left(b x-frac{b^3 x^3}{6}+frac{b^5 x^5}{120}+Oleft(x^7right) right)$$ that is to say
$$(a+2) x+left(b+frac{8}{3}right) x^3+left(frac{64}{15}-frac{b^3}{6}right)
x^5+Oleft(x^7right)$$ making $$y=frac{a+2}{x^2}+left(b+frac{8}{3}right)+left(frac{64}{15}-frac{b^3}{6}right)
x^2+Oleft(x^4right)$$ So, in order to have a limit equal to $0$, you need $a=-2$, $b=-frac{8}{3}$ and then
$$y=frac{3008 }{405}x^2+Oleft(x^4right)$$

We have that

$$frac{tan (2x)}{x^3}+frac{a}{x^2}+frac{sin bx}{x}=frac{tan (2x)+ax+x^2sin bx}{x^3}$$

and from here since $x^2sin bxsim bx^3$ 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

$$frac{tan (2x)-2x+x^2sin bx}{x^3}=8frac{tan (2x)-2x}{(2x)^3}+bfrac{sin bx}{bx}to frac83+b=0$$