Doubt at evaluating the limit for $\lim_{x \ \to \ 0} \left(\frac{\tan(2x)}{x^{3}}+\frac{a}{x^{2}}+\frac{\sin(bx)}{x} \right)=0$.

Question

Find the value of $a$ and $b$ such that,

$$\lim_{x \ \to \ 0} \left(\frac{\tan(2x)}{x^{3}}+\frac{a}{x^{2}}+\frac{\sin(bx)}{x} \right)=0$$

I noticed that I can factor an $x^{2}$ such that,

$$\lim_{x \ \to \ 0} \left(\frac{1}{x^{2}}\left(\frac{\tan(2x)}{x}+a+x\sin(bx)\right) \right)=0 $$

And by properties of limits:

$$\lim_{x \ \to \ 0}\frac{1}{x^{2}} \cdot \lim_{x \ \to \ 0} \left(\frac{\tan(2x)}{x}+a+x\sin(bx) \right)=0$$

Now since $\lim_{x \ \to \ 0} \frac{1}{x^{2}}=\infty$, it means that $\lim_{x \ \to \ 0} \left(\frac{\tan(2x)}{x}+a+x\sin(bx) \right)=0$

From there we apply L'Hospital's rule to obtain the value of $a=-2$ and $b=-\frac{8}{3}$

What I'm asking is that did I do the right thing when concluding $\lim_{x \ \to \ 0} \left(\frac{\tan(2x)}{x}+a+x\sin(bx) \right)=0$ ? or did I messed up there...

The correct argument is take limits in $\left(\frac{\tan(2x)}{x}+a+x\sin(bx) \right)=x^{2}\left(\frac{\tan(2x)}{x^{3}}+\frac a {x^{2}}+\frac {\sin(bx)}x \right)$. — Kavi Rama Murthy, Aug 08 '24 at 10:04
Is it because the fact that $\lim_{x \to 0}\frac{1}{x^{2}}=\infty$? Am I not allowed to do that? — JAB, Aug 08 '24 at 10:06
$\lim (a_nb_n)=(\lim a_n)( \lim b_n)$ if both the limits exist. You cannot use this theorem before proving that the limits exist. — Kavi Rama Murthy, Aug 08 '24 at 10:08
See the 1st limit in https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion — lab bhattacharjee, Aug 08 '24 at 10:12
I think I understood what you did, since $\lim_{x\to 0}x^{-2}=\infty$, if $\lim (\frac{\tan(2x)}x+\dots)$ is not equal to $0$, then the limit is $\infty$, so the only "hope" is to have an indeterminate form — Sine of the Time, Aug 08 '24 at 10:12
However, since you're dealing with an indeterminate form, you don't know if the limit is $0$ — Sine of the Time, Aug 08 '24 at 10:13

score 1 · Answer 1 · answered Aug 08 '24 at 10:08

The way you are using looks not so clear, we can firstly note that

$$\frac{\sin(bx)}{x}=b\frac{\sin(bx)}{bx} \to b$$

and using that $\frac{\tan x-x}{x^3}=\frac13$ since

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

taking $a=-2$

$$\frac{\tan(2x)-2x}{x^{3}}=8\cdot \frac{\tan(2x)-2x}{(2x)^{3}}=\frac 83$$

from which we can conclude that $b=-\frac83$.

heropup · Accepted Answer · 2024-08-08T10:32:40.420

No, you must not separate the factor $1/x^2$ in such a manner. To understand why, consider the following example:

$$1 = \lim_{x \to 0} \frac{\sin x}{x} \overset{?}{=} \lim_{x \to 0} \frac{1}{x^2} \lim_{x \to 0} x \sin x.$$ While the first limit tends to $\infty$, the second tends to $0$, and we have an indeterminate form. The original function, $\frac{\sin x}{x}$, actually does not tend to $0$, yet performing such a factorization as you propose would imply that the second factor, $x \sin x$, tends to $0$, which is not sufficient to establish that $\frac{\sin x}{x} \to 0$, since it actually does not.

Instead, what you should do is consider the series expansions of the trigonometric functions:

$$\tan z = z + \frac{z^3}{3} + \frac{2z^5}{15} + O(z^7),$$ which you may obtain by long division of the respective series expansions of $\sin z$ by $\cos z$. We don't actually need the $z^5$ order term, but I have included it here so that you may check your computations.

Then we have $$\frac{\tan 2x}{x^3} + \frac{a}{x^2} + \frac{\sin bx}{x} = \frac{2}{x^2} + \frac{8}{3} + \frac{a}{x^2} + b + O(x^2) = \frac{2+a}{x^2} + \frac{8}{3} + b + O(x^2).$$

In order for the limit to be $0$, the coefficients of the order $-2$ and $0$ terms must be zero; i.e., $a = -2$ and $b = -8/3$.

This clears up things for me, thanks! – JAB Aug 08 '24 at 12:58 — JAB, Aug 08 '24 at 12:58

Doubt at evaluating the limit for $\lim_{x \ \to \ 0} \left(\frac{\tan(2x)}{x^{3}}+\frac{a}{x^{2}}+\frac{\sin(bx)}{x} \right)=0$.

2 Answers2