3

The problem statement is:

Prove: If $f(x)$ is differentiable for x>0 and

$$\lim_{x \to \infty} (f(x)+\frac {df}{dx}(x)) =0$$

and if

$$\lim_{x \to \infty} f(x)$$ exists, then

$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{df}{dx}(x)=0$$

My work:

I don't see how the differentiability assumption on f can be used (or is even needed?), but I have that, for every $\epsilon$>0, there exists M = $max(M_1,M_2)$, such that x>M implies that

$$|f(x)+ \frac{df}{dx}(x)| < \epsilon$$ $$\implies-\epsilon < f(x) + \frac{df}{dx}(x) < \epsilon$$ $$\implies -\frac{df}{dx}(x) -\epsilon < f(x) < -\frac{df}{dx}(x) +\epsilon$$

and since by assumption we know that the limit at infinity for f(x) exists, then the limit must be $-\frac{df}{dx}$, by the last inequality above.

Now I think the goal is to just show that the limit at infinity of $|\frac{df}{dx}|=0.$ Then we would have the desired equality, by the Squeeze Theorem.

But what can I say about the limit of $|\frac{df}{dx}|$?

Thanks,

User001
  • 1
  • 1
    The differentiability of $f$ is already used, so that you can write $f'$. – Yes Oct 17 '15 at 06:04
  • 1
    Is $f$ required to be defined at $x = 0$? – Yes Oct 17 '15 at 06:10
  • Hi @GudsonChou, yes agreed. I thought perhaps there is another application of the differentiability of f, in order to prove this equality. What can I say about df/dx? Thanks, – User001 Oct 17 '15 at 06:10
  • Hi GudsonChou, nope - no requirements for f to be defined at x=0 ... – User001 Oct 17 '15 at 06:11
  • Hi @JohnMa, I think the linked question you suggested is not a duplicate, but rather a variation of my question. And I believe it is substantially different enough from this question, but I will read that link again just to make sure ...thanks – User001 Oct 17 '15 at 06:13
  • Hi @JohnMa, thanks again for the link - it is a pretty technique that I could not have come up with myself. I'll write it up now, and then modify my above inequalities to instead squeeze df/dx to 0, as x grows to infinity (since I will already have shown that limf(x) = 0.) – User001 Oct 17 '15 at 07:20
  • 1
    @LebronJames : If you know $\lim f(x) = 0$, then $\lim f'(x) = \lim (f'(x) + f(x) - f(x)) = \lim (f'(x) + f(x)) - \lim f(x) = 0$. But the point is how you know $\lim f(x) = 0$.... –  Oct 17 '15 at 07:40
  • Hi @JohnMa, yea, hmmm...I am a little confused with that pretty proof in the link you provided. I don't know why L'Hopital's rule can be applied; how do we know the ratio is an indeterminate form? The comments on that link mention that there is some better claim in Rudin's book that doesn't require an $\infty / \infty$ form. Is this true...? I have never heard of this before...thanks, – User001 Oct 17 '15 at 07:43
  • 1
    @LebronJames : Yes, this is less popular but is true... –  Oct 17 '15 at 07:43
  • Oh, gee ...I better learn it now. Thanks so much, @JohnMa. – User001 Oct 17 '15 at 07:45
  • Hi @JohnMa, I have not actually accepted an answer yet, but I'll re-read Gudson Chou's answer now. Can I ask you one more question, if you don't mind? Since I don't have a copy of Rudin. So, in this generalized claim of L'hopital's rule, is the criterion for the denominator to diverge to infinity or can the denominator diverge in the sense that it oscillates, say, like the cosine function? Thanks, – User001 Oct 17 '15 at 08:03
  • 1
    I just checked. The condition is that $g(x) \to \infty$ as $x\to \infty$. @LebronJames –  Oct 17 '15 at 08:04
  • Ah, ok - so divergence of the denominator to infinity, and also we are only talking about limits at infinity, and not on the unextended real line. Thanks so much for the confirmation. I really appreciate it, @JohnMa :-) – User001 Oct 17 '15 at 08:06
  • Hi @GudsonChou, sorry, I was going back and forth between the linked proof and the answers and comments on this page, too, so I didn't get a chance to see your solution in detail. But if you repost it, I'd be happy to accept it, since I think it may be the best answer. Thanks :-) – User001 Oct 17 '15 at 08:10

3 Answers3

1

With no loss of generality, suppose $\lim_{x \to \infty}f(x) > 0$; then by assumption we have $\lim_{x \to \infty}f'(x) < 0$; this implies that there are some $X \in \mathbb{R}$ and some $l < 0$ such that $f'(x) < l$ for all $x \geq X$, which in turn implies that for every $x > X$ there is some $X < c < x$ such that $$ f(x) - f(X) = (x-X)f'(c) < (x-X)l $$ by the mean-value theorem; this says that $\lim_{x \to \infty}f(x)$ does not exist, a contradiction.

Yes
  • 20,719
  • The fact that $f$ is decreasing a priori does not imply that the limit of $f(x)$ does not exists; you can say "there exists a $\lambda <0$ and $X\in \mathbb{R}$ such that $f^{\prime}(t)<\lambda$ for $t>X$, and then for $x>X$ we have $f(x)-f(X)=(x-X)f^{\prime}(c)<(x-X)\lambda$, and you are done. – Kelenner Oct 17 '15 at 07:31
  • @Kelenner Yes of course; thank you. I just found that what I typed is different from what I wrote... :) – Yes Oct 17 '15 at 07:43
1

You can force the limits into L'hopital's rule form: $$\lim_{x\rightarrow\infty}f(x) =\lim_{x\rightarrow\infty} \dfrac{e^x f(x)}{e^x}\underset{(L)}{=}\lim_{x\rightarrow\infty}\dfrac{e^x f(x) + e^x f'(x)}{e^x} = \lim_{x\rightarrow\infty}f(x)+f'(x) = 0 $$ The use of L'hopital's rule is due to the fact that all the limits exist and the 2nd from left limit is of the form of $\dfrac{*}{\infty}$ (L'hopital's rule holds for this cases aswell)
Now we can calculate the limit of the derivative:
$$\lim_{x\rightarrow\infty}f'(x) = \lim_{x\rightarrow\infty} f'(x) + f(x) - f(x) = \lim_{x\rightarrow\infty}f'(x) + f(x) - \lim_{x\rightarrow\infty} f(x) = 0-0 = 0$$

Ranc
  • 1,977
  • Hi @Ranc, for your very first equality, how do we know it is an indeterminate form, in order to apply L'hopital's rule? How does $e^xf(x)$ go to infinity? We don't know the behavior of f(x). Thanks, – User001 Oct 17 '15 at 07:38
  • 2
    @LebronJames I did not specify this, but: L'hopital rule also applies for limits of the form $\dfrac{*}{\infty}$. – Ranc Oct 17 '15 at 07:50
  • Yes, I just read up on this too. Pretty cool to know. I am guessing that this generalized L'Hopital's rule applies on all of the (extended) real line then, and not just limits at infinity, and the main criterion is just for the denominator to diverge to infinity? Thanks so much @Ranc – User001 Oct 17 '15 at 07:57
  • Would it be applicable for denominators that diverge in the sense that it just oscillates, e.g., denominator = cosine, but does not diverge to infinity? Sorry for the extra questions, but this is the first time I am encountering this generalized claim of L'Hopital. Thanks, @Ranc – User001 Oct 17 '15 at 07:58
  • 1
    @LebronJames I gave it some thought for a few minutes and: having the denominator to diverge (converge?) to $\infty$ is a somewhat easy to test. But the rule can be applied to other cases aswell. Either ways, one can always look at the reciprocal (Say, if the denominator \ numerator have zeros). Then in a way, yes, it is a main criterion.

    For the other question: Say $f$ oscillates around the limit point. What can you say about $f'$ behavior around that point? It must also oscillate (maybe in somewhat different form, but still). Meaning that also for $f'$ the limit will not exist.

     – Ranc Oct 17 '15 at 08:10
-1

If $f'(x) > M > 0$ for $x > N$, then what can you say about $f(x) = C + \int_N^x f'(x)dx$ as $x \to \infty$ ?

Paul Sinclair
  • 43,643