2

For instance we want evaluate this simple limit using taylor series :

$$L=\lim_{x\to 0}\frac{\sin x}{x^5}=\lim_{x\to 0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}$$

In this case we only care about coefficient of $x^5$ to cancel out $x^5$s in the fraction and get $L=\frac{1}{5!}$. This is a simple example of what we usually do when evaluating limit using taylor series.

My question is why we ignore other terms of the numerator? I mean:

$$L=\lim_{x\to0}\color{red}{\frac{1}{x^4}-(\frac{1}{3!}\times\frac{1}{x^2})}+\color{green}{\frac{1}{5!}}\color{red}{-(\frac{1}{7!}\times x^2)+(\frac{1}{9!}\times x^4)+\cdots}$$

I don't understand why we can ignore other terms (showed in red color) and confirm $\frac1{5!}$ as the answer. It is obvious that terms appeared on the rightside of $\color{green}{\frac1{5!}}$ are equal to zero but what about other side?

Etemon
  • 6,437
  • 1
    We can't ignore $\frac 1{x^4},\frac 1{x^3 }$ here. – PNDas Oct 26 '20 at 15:47
  • 2
    Here limit is infinite. – PNDas Oct 26 '20 at 15:48
  • @PNDas the limit is equal to $\frac1{5!}$ so we ignored all the term except green one. – Etemon Oct 26 '20 at 15:48
  • 2
    the limit is not $\frac1{5!}$ – J. W. Tanner Oct 26 '20 at 15:49
  • Sorry. I choosed a bad example. I meant in general when we use taylor series we consider only coefficient of same terms in denominator and numerator and cancel them out. for example here: https://math.stackexchange.com/questions/3881372/evaluating-lim-x-to-0-frac-cos-sin-x-1x2-frac-12x4/3881393?noredirect=1#comment8004826_3881393 we only considered coefficient of $x^4$ – Etemon Oct 26 '20 at 15:54
  • Oh. I understood it now. I thought in both case we have $x^n$ in denominator and also remain some $x^{n-k}$ on numerator. Now I understood. – Etemon Oct 26 '20 at 16:00

3 Answers3

2

We can't ignore the other terms, we obtain indeed

$$\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}=\frac1{x^4}-\frac1{3!x^2}+\frac1{5!}+O(x^2) \to \infty$$

the limit is finite for

$$\lim_{x\to 0}\frac{\sin x-\left(x-\frac{x^3}{3!}\right)}{x^5}=\lim_{x\to 0}\frac{\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+O(x^{11})}{x^5}=\frac1{5!}$$

user
  • 154,566
  • Actually, $\frac1 {x^4} - \frac 1{6x^2}$ is in $\infty - \infty$ form. You have to modify it first. – PNDas Oct 26 '20 at 15:52
  • Yes but by the standard way $$\frac1 {x^4} - \frac 1{6x^2}=\frac1{x^4}\left(1 - \frac {x^2}{6}\right) \to \infty$$ that is $1/x^4$ tends to \infty faster than all other terms. – user Oct 26 '20 at 15:53
  • You have alread answered hundreds (if not thousands) of questions about limits using Taylor series. I think it is high time you let those who have less rep show what they can. Did you even search for duplicates this time either. It is your duty as a trusted user. – Jyrki Lahtonen Oct 26 '20 at 21:14
  • 1
    @JyrkiLahtonen Sorry but I can't really understand what is your point here. This is a perfectly fine question and my answer do not prevent for other answers to be given and posted by any other user. I'll be happy to read them and in case upvote the good ones according to my preference and I'm sure that also the asker will be happy with that. Recall that we should focus our attention on the content, not the person. – user Oct 26 '20 at 21:30
  • This old thread in meta. The discussion was revitalized here and more recently here. Don't worry. I will not be targeting you. I just downvote all the high rep users when they post something that they know has already been posted. – Jyrki Lahtonen Oct 26 '20 at 21:34
2

The limit is not $\frac{1}{5!}$, in fact the limit does not exist (or, you could say the limit is $+\infty$). You need to take the limit of each individual term. Notice that the first 2 terms diverge at $x=0$ after you divide by $x^5$.

In short, you can't just blindly ignore the other terms. That only works if all the other terms go to $0$.

riemleb
  • 296
  • 1
    It is not sufficient to show that all other terms go to zero, because there can be an infinite number of them... – PierreCarre Oct 26 '20 at 16:12
0

Others have pointed out that the limit does not really exist... But, addressing the general question, it is probably easier to understand (and more correct) if you use Taylor's polynomial instead of taylor's series. Imagine you want to compute a limit that actually exists, like the well known $\lim_{x \to 0}\dfrac{\sin x}{x}$, using Taylor's formula. You know that $$ \sin x = x - \frac{\cos (\xi_x)}{3!} x^3, \quad \xi_x \in (0,x) $$

and so,

$$ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x\to 0}\dfrac{x - \dfrac{\cos (\xi_x)}{3!} x^3}{x} = 1-\frac 16 \underbrace{\lim_{x \to 0}\cos(\xi_x)x^2}_{=0} = 1 $$

PierreCarre
  • 20,974
  • 1
  • 18
  • 34