# Proof involving limits of functions and their derivatives

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.

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 $\lim_{x\to \infty} f'(x) = 0$.
• 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 Lee Sep 1 '17 at 6:11