What is the Limit of Derivative as x Approaches Infinity?

[linearfish]

Homework Statement

Suppose that f is differentiable on the interval (a, inf) and f'(x) -> 0 as x -> inf. Show that f(x)/x -> 0 as x -> inf.

The Attempt at a Solution

Conceptually I understand this problem. The derivative gets very small so as x gets large, f(x) gets very close to being a constant. Meanwhile x grows without bound so f(x)/x -> 0. However I cannot seem to get the math to work. I've tried several approaches. This seems to be my best but I'm not positive of its validity.

Fix arbitrary epsilon > 0. Then there exists an M > 0 such that | f'(x) - 0 | = |f'(x)| < epsilon.
Choose c > M and let A = f(c). Then by the MVT there exists a k in (c, inf) such that.

$$\frac{f(x + c)-f(c)}{(x+c)-c} = \frac{f(x + c)-A}{x} = f'(k) < \epsilon$$

Since A is fixed, then A/x -> 0 as x -> inf. This seems to suggest to me that we have shown the desired inequality but it bothers me I have f(x+c)/x instead of just f(x)/x. Can anyone nudge me in the right direction? Thanks.

 

[dynamicsolo]

I would propose something similar, written more along these lines:

We are given that

$$lim_{x \rightarrow \infty} f'(x) = lim_{x \rightarrow \infty} \frac{f(x) - f(a)}{x - a} = 0$$

We then perform a variable shift u = x - a , leading to

$$\frac{ lim_{u \rightarrow \infty} f(u+a) - f(0)}{lim_{u \rightarrow \infty} u} = \frac{lim_{u \rightarrow \infty} f(u)}{lim_{u \rightarrow \infty} u} - \frac{f(0)}{lim_{u \rightarrow \infty} u} = 0$$

Since f is differentiable on $$x: (a , \infty)$$ (or $$u: (0 , \infty)$$ ), f(x) must be finite (no discontinuities or vertical asymptotes there). So the second term in the difference has a limit of zero, leaving us with

$$\frac{lim_{u \rightarrow \infty} f(u)}{lim_{u \rightarrow \infty} u} = lim_{u \rightarrow \infty} \frac{f(u)}{u} = 0$$

 

[linearfish]

I follow that, thanks. Does the 'a' not really matter since u -> inf? I think that was the problem I was running into.

 

[dynamicsolo]

Since we are taking the limit as the variable approaches infiinity, the "origin" for the variable doesn't really matter (subject to the conditions for continuity and differentiability of the function, of course). It's going to be the same limit "at infinity" for f(x) regardless.

 

[Tobias Funke]

I think there's another way if you don't like this one. Only the denominator needs to go to infinity to use L'Hospital...

 

[dynamicsolo]

I think there's another way if you don't like this one. Only the denominator needs to go to infinity to use L'Hospital...

I'm afraid "L'Hopital" is not very helpful here. In this ratio, both the numerator and denominator would have to go to infinity, in order to produce the requisite indeterminate ratio $$\frac{\infty}{\infty}$$. There is no requirement in the proposition that either f(x) or f(x) - f(a) tend to infinity.

 

[Tobias Funke]

Both have to approach 0 to apply the rule to the form 0/0, but only the denominator has to approach infinity for the x/infinity case. But the only author I've seen that mentions (is aware?) of this is Rudin. I've looked at his wording so many times because I started to think I was crazy after seeing much better mathematicians than me saying you need infinity/infinity, but you don't.

 

[Dick]

I think there's another way if you don't like this one. Only the denominator needs to go to infinity to use L'Hospital...

Why do you keep saying that? lim x->infinity sin(x)/x. l'Hopital would lead to cos(x)/1. The former has a limit and the latter doesn't. Rudin may state some more generalized form, I don't doubt. But what is it? Just stating the numerator doesn't have to go to infinity is just causing confusion.

 

[Dick]

I would propose something similar, written more along these lines:

We are given that

$$lim_{x \rightarrow \infty} f'(x) = lim_{x \rightarrow \infty} \frac{f(x) - f(a)}{x - a} = 0$$

We then perform a variable shift u = x - a , leading to

$$\frac{ lim_{u \rightarrow \infty} f(u+a) - f(0)}{lim_{u \rightarrow \infty} u} = \frac{lim_{u \rightarrow \infty} f(u)}{lim_{u \rightarrow \infty} u} - \frac{f(0)}{lim_{u \rightarrow \infty} u} = 0$$

Since f is differentiable on $$x: (a , \infty)$$ (or $$u: (0 , \infty)$$ ), f(x) must be finite (no discontinuities or vertical asymptotes there). So the second term in the difference has a limit of zero, leaving us with

$$\frac{lim_{u \rightarrow \infty} f(u)}{lim_{u \rightarrow \infty} u} = lim_{u \rightarrow \infty} \frac{f(u)}{u} = 0$$

The derivative is not defined by the limit x->infinity. It's defined by the limit as a->0 first. It's a dual limit. As obvious as the correctness of the statement seems, I've given it some thought, but I still can't find a nice way to prove it. Can you help me believe you?

 

[Jaeran]

We are given that

$$lim_{x \rightarrow \infty} f'(x) = lim_{x \rightarrow \infty} \frac{f(x) - f(a)}{x - a} = 0$$

I'm a little confused. Shouldn't it start with:

$$f'(a) = lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$$

It looks like you're confusing the x in f'(x) with a separate variable you've also called x in the definition of the derivative.

Which means the question is asking you to prove:

$$lim_{a \rightarrow \infty} f'(a) = lim_{a \rightarrow \infty}lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} = lim_{a \rightarrow \infty}\frac{f(a)}{a} (somehow) = 0$$

Or with your change of variables with u = x - a:

$$lim_{a \rightarrow \infty} f'(a) = lim_{a \rightarrow \infty}lim_{u \rightarrow 0} \frac{f(u + a) - f(a)}{u} = lim_{a \rightarrow \infty}\frac{f(a)}{a} (somehow) = 0$$

I'm not sure how to follow your procedure from there, because I don't think it will work. Maybe I'm misunderstanding something you did.

 

[Tobias Funke]

Why do you keep saying that? lim x->infinity sin(x)/x. l'Hopital would lead to cos(x)/1. The former has a limit and the latter doesn't. Rudin may state some more generalized form, I don't doubt. But what is it? Just stating the numerator doesn't have to go to infinity is just causing confusion. If you just say the numerator has a finite limit, then fine, it's true. But then I didn't need l'Hopital to tell me the limit is zero.

Regarding your example, IF the resulting expression has a limit then it's the same as the original limit. If you end up with something that has no limit it doesn't mean the original has no limit. I don't mean to confuse anyone or seem as if this is all I post about, so I'll just type up the important parts of the rule (so basically all of it) as stated in baby Rudin:

Suppose f and g are real and differentiable in (a,b) and g'(x) is nonzero for all x in (a,b), where $$-\infty\leq a<b\leq \infty$$. Suppose $$\frac{f'(x)}{g'(x)}\rightarrow A$$ as x->a.

If f(x)->0 and g(x)->0 as x->a, OR if g(x)->infinity as x->a, then $$\frac{f(x)}{g(x)}\rightarrow A$$ as x->a.

In the first line of the proof it's clear that A can be +-infinity, and he remarks that the analogous statement is true if x->b or g(x)->-infinity, and as you can see there's absolutely no mention of a requirement on f if g(x)->infinity, besides the usual differentiability.

 

[Dick]

Regarding your example, IF the resulting expression has a limit then it's the same as the original limit. If you end up with something that has no limit it doesn't mean the original has no limit. I don't mean to confuse anyone or seem as if this is all I post about, so I'll just type up the important parts of the rule (so basically all of it) as stated in baby Rudin:

Suppose f and g are real and differentiable in (a,b) and g'(x) is nonzero for all x in (a,b), where $$-\infty\leq a<b\leq \infty$$. Suppose $$\frac{f'(x)}{g'(x)}\rightarrow A$$ as x->a.

If f(x)->0 and g(x)->0 as x->a, OR if g(x)->infinity as x->a, then $$\frac{f(x)}{g(x)}\rightarrow A$$ as x->a.

In the first line of the proof it's clear that A can be +-infinity, and he remarks that the analogous statement is true if x->b or g(x)->-infinity, and as you can see there's absolutely no mention of a requirement on f if g(x)->infinity, besides the usual differentiability.

Ah, ha. Thanks. I see your point. Finally. Sorry so be so dim. But if you post on this subject again, can you say f'(x)/g'(x) NEEDS to approach a limit instead of just saying the denominator doesn't have to approach infinity? That's what can cause confusion.
So I guess if I look up the proof in Rudin, It will dispel my confusion? You are probably right.

 

[dynamicsolo]

I was wondering if this wasn't somewhat hokey:

$$lim_{x \rightarrow \infty} f'(x) = lim_{x \rightarrow \infty} \frac{f(x) - f(a)}{x - a} = 0$$

What I had in mind was the limit of the slope of a secant line anchored at x = a at one end with the other end being moved off to an infinite limit. But then that doesn't tend to f'(x), does it?

Well, that was careless. I'm looking at the limit definitions for derivative (applying them correctly now), but I guess there isn't any nice way to get to the desired result. (I think about as good as that gets is that you show, with a little work, that f(x) approaches a constant limit, and thus f(x)/x approaches zero -- I'd hoped it could fall out of an application of the definition...)

linearfish may be onto a properly rigorous way to show this. As I say, I think the f(x+c) doesn't matter for the limit at infinity, but I don't know exactly what the last little step should say to bring it to the intended form.

 

[Tobias Funke]

Ah, ha. Thanks. I see your point. Finally. Sorry so be so dim. But if you post on this subject again, can you say f'(x)/g'(x) NEEDS to approach a limit instead of just saying the denominator doesn't have to approach infinity? That's what can cause confusion.
So I guess if I look up the proof in Rudin, It will dispel my confusion? You are probably right.

Yep I'll try to be clearer in the future. I think all but one of my posts has been about L'Hospital's rule lol, but usually the more advanced mathematicians take care of anything else.

I think the proof he gives should help, but it's one of his really slick ones that takes some time to go through and verify.

 

[Dick]

Yep I'll try to be clearer in the future. I think all but one of my posts has been about L'Hospital's rule lol, but usually the more advanced mathematicians take care of anything else.

I think the proof he gives should help, but it's one of his really slick ones that takes some time to go through and verify.

Oh, great. So you don't really get it either? I was hoping you could enlighten us. Could you look it up and teach us the Rudin way to the generalized l'Hopital? I don't have a copy of the book.

 

[Hurkyl]

it bothers me I have f(x+c)/x instead of just f(x)/x. Can anyone nudge me in the right direction? Thanks.

Well, the usual thing to do is to explicitly account for the difference. Consider, for example, the behavior of the expression

f(x+c)/x - f(x)/x​

or maybe

(f(x+c)/x) / (f(x)/x)​

or express the thing you have in terms of the thing you want with correction terms, such as

f(x+c)/x = (f(x) + (f(x+c) - f(x))) / x​

.

 

[Tobias Funke]

Oh, great. So you don't really get it either? I was hoping you could enlighten us. Could you look it up and teach us the Rudin way to the generalized l'Hopital? I don't have a copy of the book.

Rudin is a better teacher than I. I'm sure you can get your hands on a copy or prove it yourself, dick.

 

[Dick]

Oh, ok, Tobias. Write the MVT in the form f(x)/x=f(y)/x+(1-y/x)*f'(c) for c between x and y. Now fix a large value of y such that |f'(c)|<epsilon for all c>y. Now take x to infinity. Since y is fixed you can pick x large enough so that f(y)/x and y/x are also bounded by epsilon. This shows for large enough x, f(x)/x is bounded by a multiple of epsilon. You can generalize this to get the version of l'Hopital that Tobias was alluding to.