# Infinite derivatives of a trigonometric function

I've recently noticed that if you took an infinite amount of derivatives of a function, by that I mean something like this,

$$\lim_{x\to \infty} f^{'''\dots}(x)$$

Then if $f(x)$ is any polynomial function, then this would evaluate to either $0$, or $-\infty$, depending on whether the exponent is a fraction or not. If the function is an exponential, then the limit would be $\infty$ if the rate of exponential growth is greater than $e$, $-\infty$, if it is less that $e$, and $0$ if it is $e$.

My question:

If the function were trigonometric, then what would this expression evaluate to? All I really want are what I consider to be the three main ones: sine, cosine, and tangent.

Here's what I have so far:

Let $f(x) = \sin(x)$. $f'(x) = \cos(x),$ and $f''(x) = -\sin(x)$. I see no way to calculate a limit like this. Then I'd imagine it would go on forever like this. Do you guys have any ideas on how to calculate this limit?

• The limit does not exist. Also, The limit for $f(x)=e^x$ shoul dprobably be $e^x$. And your notation should probably be $\lim_{n\to \infty} f^{(n)}(x)$. – Hagen von Eitzen Jan 11 '14 at 22:15
• Are you sure the limit here isn't something like $\lim_{n\to\infty} f^n(x)$? Otherwise you're not just studying the infinite derivatives, rather the limit as two things go to infinity, the amount of times you derive, and the point at which you evaluate. – GPerez Jan 11 '14 at 22:16

You should define first some kind of measure, because at the very least we are talking about space of functions.. but I think you're not quite there yet, so let's just say that, intuitively, the limit does not exist.

Note this happens also with reals;

$$\lim_{x \to \infty} x \sin x$$ does not exist.

EDIT:

Let's put it this way.

Why the above limit does not exist? Because there are certain $x$ so that $\sin x > 0$, and there are other so that $\sin x < 0$. So, for very large $x$, if it happens to be $\sin x > 0$, you get close to $+\infty$, if it happens to be $\sin x < 0$, you get close to $-\infty$.

But then you come up and say "Well let's take only the $x$ so that $\sin x > 0$. This way we get arbitrarily close to $+\infty$. Indeed! Same thing for the $x$ so that $\sin x < 0$. we get close to $-\infty$ But also, let's take only the $x$ so that $\sin x = 0$. This way we get $0$.

What we did was basically considering not the function itself, but a sub sequence of it, in some sense (note that I am not being very precise, I just want to explain the concept)

So, if we can extract from a function a sub sequence that has limit $l$, than $l$ is in a limit point to $f$.

In our case the set of limit points (for $x \to \infty$) is $$\Lambda_{\infty} = \{-\infty, 0, \infty\}$$

A function $f$ has limit $l$ if and only if $\Lambda = \{l\}$.

• Could you yourself explain simply what limit points are? I did not quite understand the wikipedia article's definition. – recursive recursion Jan 11 '14 at 22:19
• @recursiverecursion for example, any irrational number is a limit point of the rational numbers. see this. As for what's a closure, think [0,1] is the closure of (0,1). Hope that helps. – GPerez Jan 11 '14 at 22:36
• @recursiverecursion hope to have cleared it a little – Ant Jan 11 '14 at 22:38
• @Ant doesn't the sub-sequence one chooses also have to go to infinity? Otherwise we could choose $x_n\to arcsin(a)$ and so $f(x_n)\to arcsin(a)a$, and get a whole other bunch of limit points. I really don't know which is why I'm asking, I was only aware of limit points of sets, not functions. – GPerez Jan 11 '14 at 22:48
• yeah well of course the sequence has to converge to the same point; that is, if we have $\lim_{x \to c} f(x)$, we have to extract a sequence so that $a_n \to c$ and use $$\lim_{n \to \infty} f(a_n)$$ :-) that is to say the set of limit points depend on the point to which x tends to, as one would expect. Tell me if something's not clear! – Ant Jan 12 '14 at 0:16

Sadly, for said trigonometric functions, the limit doesn't exists, which we can intuitively see because, continuing from where you left off, $f'''(x)=-\cos x$, and finally $f''''(x)=\sin x$, so we're back where we began. So in some way, differentiating to infinity will always be oscillating between these functions, just as evaluating $\sin x$ to infinity will always be oscillating between -1 and 1. Also, about the first couple of things you said, I think some concepts are getting mixed up. You say that you want to express taking an infinite amount of derivatives, so presumably you'd like the expression for the "infintieth" derivative. We could say that this is $$\lim_{\large\text{amount of apostrophes}\to \infty} f^{'''\dotsb}(x)$$ This is obviously sort of informal. So what we could slightly more formally write is $$\lim_{n\to \infty} f^{(n)}(x)$$

In fact, since what we want is the function resulting from the "infinitieth" derivative, we can drop the $x$, as we just want the function, not the value of it at a determined point. So an even more refined version would be: $$\lim_{n\to \infty} f^{(n)}$$ Here it's clear that the resulting limit is indeed a new function, it shouldn't be a number. There's still a few tricky parts, because recall that a limit basically says that, as $x$ is closer to $a$, $f(x)$ is closer to (some number). So here we're saying that, as $n$ is closer to infinity, $f^{(n)}$ is closer to (some function). What does it mean for two functions to be close to one another? Luckily this question has been answered, mostly, and is fundamental to what's called functional analysis, where in fact the concept of functions being infinitely close to one another is thoroughly dealt with. I don't know if you consider your question answered but I hope this was at least interesting.

• Both of you had great answers, and I did indeed find this interesting. +1, thanks for the post. – recursive recursion Jan 12 '14 at 0:14