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Suppose we have a function $f(x)$ that satisfies: $$\lim_{x\to\infty}f(x)=L$$ Where $L\in\mathbb{R}$. Is this true? $$\lim_{x\to\infty}f'(x)=0$$

My approach was simply this:

$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{xf(x)}{x}=L$$

And applying L'Hospital's rule we have:

$$\lim_{x\to\infty}\frac{xf(x)}{x}=\lim_{x\to\infty}\frac{f(x)+xf'(x)}{1}=L$$ $$\lim_{x\to\infty}f(x)+xf'(x)=L+\lim_{x\to\infty}xf'(x)=L$$ And finally: $$\lim_{x\to\infty}xf'(x)=0$$ Now, the only way this is possible is if $\lim_{x\to\infty}f'(x)\neq\infty$ and $\lim_{x\to\infty}f'(x)\neq A\in\mathbb{R}$ , because otherways the $\lim_{x\to\infty}xf'(x)$ would go to infinity. In conclusion, $\lim_{x\to\infty}f'(x)=0$

Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.

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    $\begingroup$ Something like $f(x)=\sin(x^2)/x$ provides a counterexample, doesn't it? $\endgroup$
    – John Doe
    Apr 12, 2019 at 13:33
  • $\begingroup$ To use the rule you'd need $xf$, like $x$, to diverge; but in the counterexample others have discussed herein, $xf=\sin x^2$ has no $x\to\infty$ limit. If $L$ were nonzero, on the other hand... $\endgroup$
    – J.G.
    Apr 12, 2019 at 15:24
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    $\begingroup$ Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. $\endgroup$
    – Mark Viola
    Apr 12, 2019 at 16:45
  • $\begingroup$ @MarkViola: On the other hand, then you do need the denominator to approach $\infty$. Otherwise you get into trouble with cases like $\lim\limits_{x\to\infty} \frac{2-1/x}{1-1/x}$. $\endgroup$
    – hmakholm left over Monica
    Apr 12, 2019 at 20:50
  • $\begingroup$ @HenningMakholm Indeed. The limit of the denominator must approach $\infyt$ (or $-\infty$). $\endgroup$
    – Mark Viola
    Apr 12, 2019 at 22:03

2 Answers 2

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Suppose that $f(x)=\dfrac{\sin(x^2)}x$. Then $\lim_{x\to\infty}f(x)=0$, but the limit $\lim_{x\to\infty}f'(x)$ doesn't exist.

If you try to apply L'Hopital's Rule here as you did, you will be working with$$\lim_{x\to\infty}\frac{x\sin(x^2)}{x^2}.$$But if $g(x)=x\sin(x^2)$, then the limit $\lim_{x\to\infty}g'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.

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    $\begingroup$ so the only thing we can conclude is: if $\lim_{x \to \infty} f'(x)$ exists, then it must be $0$? $\endgroup$
    – antkam
    Apr 12, 2019 at 13:39
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    $\begingroup$ @antkam Yes, that is correct. $\endgroup$
    – José Carlos Santos
    Apr 12, 2019 at 14:10
  • $\begingroup$ Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. $\endgroup$
    – Mark Viola
    Apr 12, 2019 at 16:45
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    $\begingroup$ @marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = \sqrt{1} = \sqrt{-1·-1} = \sqrt{-1}·\sqrt{-1} = i·i = -1$. If all rules were taught properly, such fallacies would never even arise. $\endgroup$
    – user21820
    Apr 13, 2019 at 3:09
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(Paraphrased from Wikipedia.)

L'Hôpital's rule:

Given functions $f$ and $g$ which are differentiable on an open interval $I$, except possibly at a point $c \in I$, if

$$ \lim _{x \to c}F(x)=\lim _{x\to c}G(x)=0 \text{ or }\pm \infty, \tag{1.} $$ $$ G'(x)\neq 0 \text{ for all }x \in I, \text{ with }x \ne c, \text{ and} \tag{2.} $$ $$ \lim_{x \to c}\frac{F'(x)}{G'(x)} \text{ exists.} \tag{3.} $$

then

$$\lim_{x \to c} \frac{F(x)}{G(x)} =\lim_{x \to c} \frac{F'(x)}{G'(x)}. \tag{4.}$$

You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, \infty)$ for some $x_0 < 0$.

Since $\lim _{x\to \infty}G(x)= \infty$, condition $(1.)$ requires that $$\lim _{x \to \infty}xf(x) = \infty. \tag{A.}$$

Condition $(2.)$ is satisfied by $G(x)=x$.

Condition $(3.)$ requires that $$\lim_{x \to \infty}[f(x)+xf'(x)] \text{ exists.} \tag{B.}$$

If conditions $(A.)$ and $(B.)$ are met, then, by L'Hôpital's rule, $$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty}[f(x)+xf'(x)]$$

Others have shown you that counter examples do exists.

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  • $\begingroup$ Application of LHR does not require that the numerator approach $\infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit. $\endgroup$
    – Mark Viola
    Apr 12, 2019 at 16:45
  • $\begingroup$ First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $\lim F=\infty$. $\endgroup$
    – Mark Viola
    Apr 12, 2019 at 22:08
  • $\begingroup$ Is that all you ascertained from my note? Read the second sentence of that note and revert if you have anything to add. $\endgroup$
    – Mark Viola
    Nov 9, 2021 at 22:38

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