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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!

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  • $\begingroup$ Taylor expand $tan(2x)$ perhaps? $\endgroup$ – More Anonymous Nov 26 '18 at 1:12
  • $\begingroup$ Have you tried writing it as one fraction and applying Lhopotal's? The denominator is $x^3$ so it should become a constant after several iterations. $\endgroup$ – Ovi Nov 26 '18 at 1:15
  • $\begingroup$ How does $\tan(2x)/x^3$ "approaches infinity as x goes to infinity" since it's not even defined on any neighborhood of $\infty$? You meant "as $x$ appraoches zero," right? $\endgroup$ – Clement C. Nov 26 '18 at 1:18
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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 $b=-8/3$.

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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 $a=-2$, $b=-\tfrac83$.

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  • $\begingroup$ The correct answer is -2 for a and -8/3 for b $\endgroup$ – Dude156 Nov 26 '18 at 1:44
  • $\begingroup$ Yeah, I forgot the 2 from the $\tan 2x$. $\endgroup$ – Martin Argerami Nov 26 '18 at 2:11
  • $\begingroup$ Thanks for the reply! Why couldn't we solve this by solving each limit independently? Also, why did you say that if a=0, we couldn't use l hospitals? $\endgroup$ – Dude156 Nov 26 '18 at 2:20
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    $\begingroup$ Yes, that was not entirely accurate. I meant that you cannot guarantee that the limit of the quotient of the derivatives exists. The problem with L'Hôpital is that when the limit of $f'(x)/g'(x)$ doesn't exist, it doesn't tell you anything about the existence of the original limit. $\endgroup$ – Martin Argerami Nov 26 '18 at 2:24
  • $\begingroup$ Hey what does this o notation stand for btw? How did you dodge sec^2(x)? $\endgroup$ – Dude156 Nov 26 '18 at 17:36
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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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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

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

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