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Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function satisfying $\lim_{x \to \infty} ( f(x) + f'(x) ) = 0.$ Prove that $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} f'(x) = 0.$

I know this is a copy of this question, but I don't want to use L'Hopital because my teacher doesn't allow it. I tried the following, is this correct?

Let $h(x) := f(x)e^x$. Then $h'(x) = f'(x) e^x + e^x f(x)$. Thus

$h'(x) = e^x (f'(x) + f(x)).$

We know $\lim_{x \to \infty} ( f(x) + f'(x) ) = 0$, and so

$\Rightarrow |f'(x) + f(x)| < \epsilon$
$\Rightarrow -\epsilon < f' + f < \epsilon$
$\Rightarrow - \epsilon e^x < e^x (f' + f) < \epsilon e^x$
$\Rightarrow -\epsilon e^x < h'(x) < \epsilon e^x$
$\Rightarrow \int_{x_0}^{x} h'(t) \, dt < \epsilon \int_{x_0}^{x} e^t \, dt$
$h(x) - h(x_0) < \epsilon (e^x - e^{x_0}).$

Dividing by $e^x$, we see

$$\frac{h(x) - h(x_0)}{e^x} < \epsilon \left( 1 - \frac{e^{x_0}}{e^x} \right).$$

Since $\lim_{x \to \infty} \frac{e^{x_0}}{e^x} = 0$, we get $f(x) < \epsilon$. Similarly, we can show that $-\epsilon < f(x)$. Thus, by the Squeeze Theorem, $\lim_{x \to \infty} f(x) = 0$, and from this $\lim_{x \to \infty} f'(x) = 0$.

jpmacmanus
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Leviee
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  • Why does your teacher not allow use of L'hopital's Rule? – K41 Feb 24 '25 at 19:25
  • idk sir ask aye ess aye – Leviee Feb 24 '25 at 19:46
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    Here are few answers which have solved this same question without LH 1. https://math.stackexchange.com/q/648652/1243299 2. https://math.stackexchange.com/a/407692/1243299 – Aditya Teraiya Feb 24 '25 at 19:50
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    From https://math.stackexchange.com/tags/solution-verification/info: "Is my proof correct?" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof is in doubt, and why so. – Martin R Feb 24 '25 at 19:58
  • @MartinR I'm specifically talking about the last step where I took the limit? and if you identify any other flaw except that i'd be glad to know – Leviee Feb 24 '25 at 20:07
  • @AdityaTeraiya thanks the second solution looks similar to mine :D – Leviee Feb 24 '25 at 20:10
  • The usual reason for disallowing L'Hopital's rule is that students tend to overly rely on it, and thus never properly develop their ability to use other tools, such as the definition of a limit and the squeeze theorem. Coming up with sample problems for which L'Hopital's rule is not applicable generally requires functions that neophytes are not comfortable with using, making the problems harder for them. So using problems where the rule is applicable, but not allowing it, provides simpler problems for them to work out how to use the other tools. – Paul Sinclair Feb 27 '25 at 19:17
  • The only problem I see with your approach is that it assumes integrability of $h(x) $ which is equivalent to that of $f(x) $. Use mean value theorem and mimic the proof for L'Hospital Rule. – Paramanand Singh Feb 28 '25 at 06:58
  • You may also have a look at this old answer of mine. – Paramanand Singh Feb 28 '25 at 07:05

1 Answers1

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l'Hopital's Rule is basically Cauchy Mean Value Theorem in disguise, so you can in turn use Cauchy Mean Value Theorem instead.

Let $\varepsilon > 0$. Since $\displaystyle\lim_{x\to\infty} |f(x) + f'(x)| = 0$, by definition, there exists $N>0$ such that $$ x > N \implies |f(x) - f'(x)| < \dfrac{\varepsilon}{2}. \tag{$\star$} $$

Your intuition for $h(x) = e^x f(x)$ is great. Suppose indeed $$ x > \max\left\{N, \frac{2e^N|f(N)|}{\varepsilon}\right\} > 0, $$

we can apply Cauchy Mean Value Theorem to $h(x)$ and $e^x$ on $[N,x]$ to get some $\xi\in(N,x)$ such that $$ \frac{e^x f(x) - e^N f(N)}{e^x - e^N} = \dfrac{e^\xi(f(\xi) + f'(\xi))}{e^\xi} = f(\xi) + f'(\xi). $$ Multiply both sides by $\dfrac{e^x - e^N}{e^x} = 1 - \dfrac{e^N}{e^x}$ and taking absolute values, we have $$ \left|f(x) - \frac{e^N}{e^x}f(N)\right| = \left|(f(\xi) + f'(\xi)) \left(1 - \frac{e^N}{e^x}\right)\right| \overset{(\star)}{<} \frac{\varepsilon}{2}\left(1 - \frac{e^N}{e^x}\right). $$ By a simple use of triangle inequality, we get $$ |f(x)| < \left|\frac{e^N}{e^x}f(N)\right| + \frac{\varepsilon}{2}\left(1 - \frac{e^N}{e^x}\right) \leq \left|\frac{e^N}{e^x}f(N)\right| + \frac{\varepsilon}{2} $$

Similar to your proof, I think your concern is you wish to take limit of $x$ in the terms with $\frac{1}{e^x}$ only, to do this rigorously, I have set $x > \frac{2e^N|f(N)|}{\varepsilon}$ from above. Then we see $$ |f(x)| \overset{\frac{1}{e^x} \leq \frac{1}{x} < \frac{\varepsilon}{2e^N|f(N)|}}{<} \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$ By definition, we conclude $\displaystyle\lim_{x\to\infty} |f(x)| = 0$.

Samuel Ho
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