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I want to prove or disprove this problem: If there exist $\lim\limits_{x\rightarrow \infty} (f'(x)+f(x))=L<\infty$ then $\lim\limits_{x\rightarrow\infty} f(x) =L$.

When I assume problem below:

If there exist $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, There exists $\lim\limits_{x\rightarrow\infty} f(x)$?

I can use mean-value theorem to show that.

So my question is:

If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x))=L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x)$ exist?

Michael Hardy
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Maddy
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    Why did you include the same phrase thrice in your question? Couldn't everything after the first sentence be dropped without changing the meaning? (Also, you can use \lim in math mode to get $\lim$, and \lim\limits_{x\to\infty} gives $\lim\limits_{x\to\infty}$.) – Lord_Farin May 31 '13 at 12:27
  • @julien: I had to read it thrice, but there was a subtle difference. He was alleging that he knew how to prove that $\lim_{x\to\infty} f(x)$ exists in one of the statements. He was then asking how to show that limit must be $L$. – Ted Shifrin May 31 '13 at 14:11
  • @TedShifrin Good point. I read too quickly, thanks. – Julien May 31 '13 at 14:13

2 Answers2

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Consider the function $$g(x)=e^{x} f(x).$$ Then $$Dg(x)=e^{x}f(x)+e^{x}Df(x)=e^{x} \left( f(x)+Df(x) \right).$$ Now, $$ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{g(x)}{e^x} = \lim_{x \to +\infty} \frac{Dg(x)}{e^x} = \lim_{x \to +\infty} Df(x)+f(x) $$ by De l'Hospital's theorem.

N.B. I think this exercise was solved by G. Hardy in one of his books.

Siminore
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  • Beautiful Proof. Thanks. – Maddy May 31 '13 at 13:33
  • It's not mine, of course. – Siminore May 31 '13 at 13:34
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    To apply l'Hôspital's rule, is it not needed that $\lim\limits_{x \to + \infty} g(x)=+ \infty$? – Seirios May 31 '13 at 13:38
  • @Seirios I know that $\lim(e^x)=\infty$ is sufficient. – Maddy May 31 '13 at 13:44
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    @Maddy No that's not sufficient to get $\lim g=+\infty$. Not every $f$ has $\lim_{+\infty}e^xf(x)=+\infty$. – Julien May 31 '13 at 13:47
  • So I am curious as to how we have $ \lim g = \infty $?? – Vishesh May 31 '13 at 13:59
  • @Vishesh We have, unless $L=0$ – Hagen von Eitzen May 31 '13 at 14:11
  • What happens to this proof when $f$ is of exponential decay (to zero or otherwise)? I have a feeling that the L'Hospital step breaks down since $g$ may not go to $\pm \infty $. – Cameron Williams May 31 '13 at 14:11
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    In fact, we can suppose that $L>0$ (otherwise, consider $f+n$ for $n$ large enough). So there exists $a>0$ such that $g'(x)>a$ for $x$ large enough. If $g'$ is continuous, it is easy to deduce that $g(x)>a(x-x_0)$ and so $g(x) \to + \infty$. Otherwise, I don't know... – Seirios May 31 '13 at 14:13
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    @CameronWilliams: In that case, $\lim_{x\to\infty} f(x) = 0$ anyway (or to $L$, where $L$ is the limit of the exponential decay). The proof above strictly proves the contrapositive - that if the limit of $f(x)$ doesn't exist, then neither does the limit of $f(x)+f'(x)$. – Glen O May 31 '13 at 14:13
  • I second that. Thats what I feel too – Vishesh May 31 '13 at 14:15
  • To apply De l'Hospital's theorem, you just need the obvious fact that $\lim_{x \to +\infty} e^x=+\infty$. – Siminore May 31 '13 at 14:22
  • @julien What missing argument? – Siminore May 31 '13 at 14:27
  • @julien Definitely no: when the denominator diverges, you should not care of what the numerator does. It is the form [anything/$\infty$]. – Siminore May 31 '13 at 14:37
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    @julien Pick up Walter Rudin's Priciples of mathematical analysis. You'll read that "if either $f(x) \to 0$ and $g(x) \to 0$, or if $|g(x)| \to +\infty$, then..." I agree that the most interesting case of De l'Hospital's theorem concerns the case $[\infty/\infty]$, but it *is true when the denominator diverges, no matter what the numerator does. Please consider that Wikipedia is not the best source for optimal mathematical statements :-) – Siminore May 31 '13 at 15:04
  • Oh, that statement, I see what you mean, now. We were all commenting on the basis of the wikipedia statement. And I only wanted to point out that when $g'$ and $h'$ tend to $+\infty$, automatically $g$ and $h$ do as well, which is the case here, and does not require to look at $g$ and $h$ at all. Of course, Rudin's statement is better than wikipedia's. – Julien May 31 '13 at 15:18
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    What a wonderful proof. Do you have a more specific citation? – Gamma Function Jun 07 '13 at 10:10
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    The statement, with a rather old-fashioned hint, appears as Exercise 32, page 246, of Hardy, A course of pure mathematics, Cambridge University Press, 1908. I am pretty sure I studied this proof on another book, but I can't remember what book right now. – Siminore Jun 07 '13 at 11:34
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When $f'$ is continuous, you can avoid l'Hôspital's rule by saying:

Let $\epsilon >0$. There exists $x_0 \in \mathbb{R}$ such that $x \geq x_0$ implies $$L-\epsilon < f'(x)+f(x) <L+\epsilon$$

Hence $$(L-\epsilon)e^x < (f'(x)+f(x))e^x <(L+\epsilon)e^x$$ $$\int_{x_0}^t(L-\epsilon)e^xdx < \int_{x_0}^t (f'(x)+f(x))e^xdx <\int_{x_0}^t (L+\epsilon)e^xdx$$ $$(L-\epsilon)e^t-(L-\epsilon)e^{x_0} < e^tf(t)-f(0)<(L+\epsilon)e^t-(L+ \epsilon)e^{x_0}$$ $$(L-\epsilon) - (L-\epsilon)e^{x_0-t} < f(t)-f(0)e^{-t} < (L+ \epsilon)-(L+\epsilon)e^{x_0-t}$$

for $t \geq x_0$. When $t$ is large enough, we get $$L-2\epsilon < f(t) < L+2\epsilon$$

Therefore, $\lim\limits_{x \to + \infty} f(x)=L$.

Michael Hardy
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Seirios
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