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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?

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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. – 1015 May 31 '13 at 14:13
up vote 9 down vote accepted

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.

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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
@Maddy No that's not sufficient to get $\lim g=+\infty$. Not every $f$ has $\lim_{+\infty}e^xf(x)=+\infty$. – 1015 May 31 '13 at 13:47
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
@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
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 7 '13 at 11:34

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

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Why $(f'(x)+f(x))e^x$ is integrable? – Maddy May 31 '13 at 13:53
Indeed, I added the assumption that $f'$ is continuous in my answer. – Seirios May 31 '13 at 14:02
@ Seirios.There are discontinuous functions that are integrable too right??Doesnt this restrict the scope of the proof?? – Vishesh May 31 '13 at 14:20

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