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Suppose that $f: \mathbb{R^+}\to \mathbb{R}$ satisfies $\lim_{x\to \infty} (f+f')(x)=0$. Show that $\lim_{x\to \infty} f(x)=0$.

This is one solution I found to this problem.

Solution: If $x=a$ is in the domain of the function $f$, then by the generalized mean value theorem applied to the functions $e^x f(x)$ and $e^x$, there exists $c\in (a,x)$ such that $\frac{e^x f(x)-e^a f(a)}{e^x -e^a}=\frac{e^c f'(c)+e^c f(c)}{e^c}.$ This yields $\frac{e^x}{e^x -e^a} f(x)-\frac{e^a}{e^x - e^a}f(a)=f'(c)+f(c)$. Taking limits as $c$ tends to infinity we have, $\lim_{c\to \infty}\frac{e^x}{e^x -e^a} \lim_{c\to \infty}f(x)-\lim_{c\to \infty}\frac{e^a}{e^x - e^a}\lim_{c\to \infty}f(a)=\lim_{c\to \infty}[f'(c)+f(c)]$. But $c\lt x$, thus we obtain $1 \cdot \lim_{x\to \infty}f(x)-0\cdot f(a)=\lim_{c\to \infty}[f'(c)+f(c)].$ Hence, since the right hand side is equal to $0$, it follows that $\lim_{x\to \infty}f(x)=0.$

Taking a close look at this solution, however, I think it may not be correct since the $c$ used in the answer is actually a function of $x$ and it may not make sense to take $c\to \infty$. I mean how do we know that there will always be such a $c$ greater than any given positive number? Is this solution correct? If so, how can my question be answered? I would greatly appreciate any help.

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  • $\begingroup$ how about spliting the limit $\lim_{x \to \infty}{f}=-\lim_{x \to \infty}{f'}$, now suppose that $\lim_{x \to \infty}{f}=c < \infty$, by continuity of $f$ and mean-value theorem this would have to be zero $\endgroup$
    – Nikos M.
    Sep 12, 2015 at 0:53
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    $\begingroup$ But in order to split the limit we need to know that the each limit exists. $\endgroup$ Sep 12, 2015 at 0:55
  • $\begingroup$ In the first line of your solution, I assume you mean $x > a$? $\endgroup$
    – Clement C.
    Sep 12, 2015 at 0:57
  • $\begingroup$ @ClementC. That just means that we fix some random positive $a$ is in the domain. $\endgroup$ Sep 12, 2015 at 0:58
  • $\begingroup$ Then, you cannot really use later $x$ anymore: technically, you have a conflict of notation (you "binded" $x$ to $a$ by writing "$x=a$"). $\endgroup$
    – Clement C.
    Sep 12, 2015 at 1:00

3 Answers 3

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Let $g(x) =f(x)+f'(x) $ so $g(x) \to 0 $ as $x \to \infty $.

Then, for any $c > 0$, there is a $d(c)$ such that $|g(x)| < c$ for $x > d(c) = d$.

Then $|e^x g(x)| < ce^x $ for $x > d$.

But $e^xg(x) =e^x(f(x)+f'(x)) =(f(x)e^x)' $ so, for $x > d$, $f(x)e^x-f(d)e^d =\int_d^x e^t g(t) dt $.

Therefore, for $x > d$,

$\begin{array}\\ |f(x)e^x| &=|f(d)e^d + \int_d^x e^t g(t) dt|\\ &\le |f(d)e^d| + |\int_d^x e^t g(t) dt|\\ &\le |f(d)e^d| + |\int_d^x e^t c dt|\\ &= |f(d)e^d| + c(e^x-e^d)\\ \text{or}\\ |f(x)| &\le |f(d)e^{d-x}| + c(1-e^{d-x})\\ &< |f(d)e^{d-x}| + c\\ \end{array} $

and the right side can be made arbitrarily small by first, making $c$ small and then, making $x$ large.

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To fix the problem, instead of arbitrary $ a < x$ for a resulting $c\in(x,2x)$, consider $x$ (instead of your constant $a$) and $2x$ (instead of your $x$); i.e., $(x,2x)$. Then you get $c=c_x\in(x,2x)$, meaning that you can make $x$ go to infinity: this will imply $c\to\infty$ as well.

Edit: only fixes one issue in the proof -- the second half then fails, and this answer as it stands is not a good answer. There may be a way to go through with the same approach as your original one, by considering $x$ and $x+t$, for $x,t > 0$. By the same argument as yours and above, you get $c=c(x,t)$; the question is now whether you can make $x$ and $t$ go to infinity "in a decoupled fashion" for everything to go through. (It seems very messy to me, while possibly doable; but I'd recommend the "differential-equation" approach in the first comment below, instead.)

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  • $\begingroup$ For completely different, alternative, solution: let $g = f+f^\prime$ (it's a function). Solve the differential equation $f^\prime + f = g$ (the solution will be a function of $g$). Now, using the only thing you know about $g$ -- namely, that its limit at $\infty$ is zero, use the expression of $f$ you found for the solution to the differential equation to show that it goes to zero as well. $\endgroup$
    – Clement C.
    Sep 12, 2015 at 1:07
  • $\begingroup$ But if I fix the problem as you advised, the solution does not work out since $\lim_{c\to \infty} \frac{e^a}{e^x -e^a}=-1$ not $0$. $\endgroup$ Sep 12, 2015 at 1:11
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    $\begingroup$ In the above, I suggested $c\in(x,2x)$, which with your notations corresponds to "$a=x$, and "$x=2x$." (Sorry, I had inverted the order between your $x$ and your $a$, which brings some confusion -- I edited my answer to be more consistent with your original notations). You do have both $\frac{e^{2x}}{e^{2x}-e^x} \to 1$ and $\frac{e^{x}}{e^{2x}-e^x} \to 0$ (when $x\to\infty$), as wished (unless I missed something reading the structure of your argument) $\endgroup$
    – Clement C.
    Sep 12, 2015 at 1:17
  • $\begingroup$ But in order for $\lim_{c\to \infty}\frac{e^{2x}}{e^{2x} -e^a} f(x)-\frac{e^a}{e^{2x} - e^a}f(a)$ to be equal to $\lim_{c\to \infty}\frac{e^{2x}}{e^{2x} -e^a} \lim_{c\to \infty}f(2x)-\lim_{c\to \infty}\frac{e^a}{e^{2x} - e^a}\lim_{c\to \infty}f(a)$, we need to know that the limit of $f(x)$ exists, so that the multiplication splits. Also since using your notation, $a$ is no longer fixed, $\lim_{c\to \infty} f(a)\neq f(a)$. I'm having a feeling that the solution is just wrong, can it still be fixed? $\endgroup$ Sep 12, 2015 at 1:28
  • $\begingroup$ I am going through he second part of the proof -- there is indeed an issue with what you mention (namely, the second term does not nicely become $0\cdot\lim f$). There should be a way to salvage it, by decoupling the two bounds ($a$ and $x$) nicely and making them to go to infinity separately... I do believe, however, that the most direct way to solve the exercise is with the differential equation approach (see comment above). $\endgroup$
    – Clement C.
    Sep 12, 2015 at 1:31
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Let $g(x)=f(x)e^x$. Then $$ g'(x)=(f(x)+f'(x))e^x $$ By L'Hospital's rule, we have $$ \lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{g(x)}{e^x}=\lim_{x\to\infty}\frac{g'(x)}{e^x}=\lim_{x\to\infty}(f(x)+f'(x))=0 $$

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  • $\begingroup$ This is the shortest possible answer to the question. +1 $\endgroup$
    – Paramanand Singh
    Sep 12, 2015 at 5:38
  • $\begingroup$ Unfortunately, this isn't valid. To use L'Hospital's rule here, one requires $\lim_{x\to\infty} g(x) = \infty$, which may not be true (e.g. if $f \equiv 0$). $\endgroup$ Sep 12, 2015 at 22:45
  • $\begingroup$ if $f≡0$, then $g≡0$. $\endgroup$
    – hermes
    Sep 12, 2015 at 23:00
  • $\begingroup$ @SameerKailasa: For LHR one only needs the denominator to tend to $\infty$. $\endgroup$
    – Paramanand Singh
    Sep 13, 2015 at 3:06

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