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I'm working on another analysis problem: Let $f$ be a differentiable function on an interval of the form $(a,+\infty)$. Prove that if there is a number $r > 0$ such that $\lim_{x\to\infty}(rf′(x)+f(x))=L$ is finite, then $\lim_{x\to\infty}f′(x)=0$ and $\lim_{x\to\infty}f(x)=L$.

I figured out the proof in the case where $r>0$ and $\lim_{x\to\infty}f(x)$ exists and is nonzero and the case where $r<0$ and $\lim_{x\to\infty}f(x)$ exists and is finite (it's a pretty simple application of L'Hôpital's Rule, once you establish it applies). However, I can't seem to be able to show that $\lim_{x\to\infty}f(x)$ must exist. I also don't know where to go with the zero case for $r>0$ and the infinite case for $r<0$.

If it helps, this is in the section on L'Hôpital's Rule in the chapter on derivatives in Joseph L Taylor's Foundations of Analysis. Any help will be well appreciated.

Michael L.
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themathandlanguagetutor
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1 Answers1

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There is a simpler answer that comes from a well-documented trick. See that \begin{align*} \lim_{x\to \infty} f(x) &= \lim_{x\to \infty} \frac{e^{x/r}f(x)}{e^{x/r}} \\ &= \lim_{x\to \infty} \frac{e^{x/r}(f'(x)+f(x)/r)}{e^{x/r}/r}\text{ (l'Hôpital)} \\ &= \lim_{x\to \infty} \frac{e^{x/r}(rf'(x)+f(x))}{e^{x/r}} \\ &= \lim_{x\to \infty} [rf'(x)+f(x)] \end{align*} This trick also works in the solution that uses differential equations, but it is done implicitly here. There is a chart here that justifies the use of l'Hôpital.

To solve this problem using differential equations theory, write $$rf'(x)+f(x) = L+\epsilon(x)$$ where $\lvert \epsilon(x)\rvert$ is vanishingly small as $x\to \infty$. The solution to this ODE with initial condition $f(x_0) = c$ is $$f(x) = e^{-x/r}\left(ce^{x_0/r}+\int_{x_0}^x \frac{L+\epsilon(\xi)}{r}e^{\xi/r}\,\mathrm{d}\xi\right)$$ Then, l'Hôpital and the fundamental theorem of calculus give us $$\lim_{x\to \infty} f(x) = \lim_{x\to \infty} \frac{ce^{x_0/r}+\int_{x_0}^x \frac{L+\epsilon(\xi)}{r}e^{\xi/r}\,\mathrm{d}\xi}{e^{x/r}} = \lim_{x\to \infty} \frac{\frac{L+\epsilon(x)}{r}e^{x/r}}{e^{x/r}/r} = \lim_{x\to \infty} [L+\epsilon(x)] = L$$ From this, we can conclude that $\lim_{x\to \infty} f'(x) = 0$.

Michael L.
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  • The problem I have with this approach is that, since this is an introductory analysis course, we shouldn't be using even calculus unless it's a result that's already been proven earlier in the course, let alone differential equations. – themathandlanguagetutor Sep 01 '17 at 06:01
  • In addition, even just showing that the function satisfies the differential equation, I don't think I'd have the tools to show that L'Hôpital's Rule applies. – themathandlanguagetutor Sep 01 '17 at 06:07
  • It's not hard to show that either both the numerator and denominator have infinite limits or both go to $0$ (depending on the sign of $r$), although it's a bit of a moot point if this solution is too far removed from the intended method. How much calculus do you have at this point? Can you use the fundamental theorem? – Michael L. Sep 01 '17 at 06:11
  • pretty slick proof though--not gonna lie. – Rustyn Sep 01 '17 at 06:16
  • We are allowed to use derivatives of elementary functions and the fact that they're continuous on their natural domains. We've defined limits (including one-sided and infinite) derivatives, and we've established the basic rules of how to use them (such as linearity, product rule, etc.) We've also established the Mean Value Theorem and Cauchy's Mean Value Theorem, as will as several involving continuity and uniform continuity. We also have quite a few theorems regarding sequences. What we do not yet have is integrals; we haven't even defined them (in this course). – themathandlanguagetutor Sep 01 '17 at 06:18
  • That said, there are times where I'm a little unclear about how much calculus I'm actually able to take for granted. – themathandlanguagetutor Sep 01 '17 at 06:19
  • @rustyn yes indeed. It would make my life much easier if I could just use this approach haha – themathandlanguagetutor Sep 01 '17 at 06:51
  • Done! There's actually a very nice trick that produces the answer extremely easily. – Michael L. Sep 01 '17 at 08:16
  • Although, the justification of l'Hôpital in the pure analysis solution is messier than the justification of l'Hôpital in the ODE solution. – Michael L. Sep 01 '17 at 08:30
  • The application of L'Hospital's Rule in case of $r<0$ does not appear valid as the denominator tends to $0$, but the numerator does not. – Paramanand Singh Sep 01 '17 at 19:05
  • @Michael Lee Thank you! Upon looking through the sources you have and the sources there and so on, I discovered that I didn't look through the proof of L'Hôpital's Rule presented in the book closely enough, because although the theorem is presented with the hypothesis that both limits are infinite, only the bottom needs to be! – themathandlanguagetutor Sep 01 '17 at 22:39
  • As for the case where r<0, it turns out that the result is not true (the book just miswrote the theorem, not the first time it's happened) so @Paramanand Singh that would explain why it's not valid. That said, the numerator actually would tend to zero, provided f had a finite limit – themathandlanguagetutor Sep 01 '17 at 22:42
  • I fixed the question to account for the typo. There's an easy counterexample for all $r < 0$, which is that $rf'(x)+f(x) = L$ is solved by $f(x) = ke^{-x/r}+L$ for any $k$, which clearly diverges as $x\to \infty$ for $k\neq 0$. – Michael L. Sep 01 '17 at 22:44
  • +1 (given already) for this excellent answer. The treatment via differential equations is important. See the following generalization : https://math.stackexchange.com/q/726027/72031 – Paramanand Singh Sep 02 '17 at 02:06