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I have come up with a theorem. (It probably doesn't qualify as a theorem, but I don't know what else to call it) This is what it states:

If $$\lim_{x\to\infty}f'(x)$$ exists (is finite), then $$\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}f'(x)$$ I have yet to find a counterexample, and it just seems to make sense to me, but I can't actually prove it.

This "theorem" is crucial to an answer to a question that I was asked to give, but my answer won't be a very good one if I can't prove why it works.

What I've tried:

I have been able to prove it if $f(x)$ has a linear asymptote (I think my terminology is correct).

If:

$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}mx+b$$

Then:

$$\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}\frac{mx+b}x=m$$

But that doesn't work if there is no asymptote. Take $f(x)=\ln(x)$ for example. the $b$ in $mx+b$ would be infinite.

Polygon
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    Sounds like a case of de l'hopitals theorem? – Myself Aug 17 '16 at 19:17
  • Note the limit to infinity of f is the average value of f truncated to domain (0, infinity) . – Jacob Wakem Aug 17 '16 at 20:05
  • @Alephnull do you mean f'? Also, I think that would only work if f is continuous for all positive numbers. That can be fixed, though, by changing 0 to some number larger than the biggest value where f is discontinuous. – Polygon Aug 17 '16 at 20:50
  • @Polygon it is general for all functions such as sequences. For each epsilon, almost all f (x) are in lim f (x)+/- epsilon. (All but a finite length or all but finitely many). It works better for integrable functions,because you can take the running average. – Jacob Wakem Aug 17 '16 at 20:58

3 Answers3

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A rigorous way to show this is by invoking $\varepsilon$-$M$ arguments. By condition $c \equiv \lim_{x \to \infty} f'(x)$ exists and finite, which implies that for any given $\varepsilon > 0$, there exists $M > 0$ such that for all $x \geq M$, $$|f'(x) - c| < \varepsilon. \tag{1}$$ On the other hand, by the mean value theorem, for any $x > M$, we have $$f(x) = f(M) + f'(\xi)(x - M)$$ for some $\xi \in (M, x)$. Therefore, for all $x > M$, it follows that \begin{align*} & \left|\frac{f(x)}{x} - c\right| \\ = & \left|\frac{f(M) + f'(\xi)(x - M)}{x} - c\right| \\ = & \left|f'(\xi) - c + \frac{f(M) - f'(\xi)M}{x}\right| \\ \leq & |f'(\xi) - c| + \frac{|f(M) - f'(\xi)M|}{x} \\ < & \varepsilon + \varepsilon = 2\varepsilon. \end{align*} In the last step we may increase $x$ whenever necessary so that the latter term is bounded by $\varepsilon$, it is possible to do so since the numerator is bounded by a fixed number. This completes the proof.

Zhanxiong
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L'Hopital's Rule does not require the numerator to approach infinity. In fact, the limit $\lim_{x\to \infty}f(x)$ need not even exist. See the note HERE that references the case of interest.

Thus, if $f'(x)$ exists for some interval $(x_0,\infty)$ and $\lim_{x\to \infty}\frac{f'(x)}{1}$ exists, then we have

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

Mark Viola
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  • Nice answer. I have read the Wikipedia page, but can you confirm: the version of L'Hopital's Rule you are using requires that (i) $\frac{f'(x)}{g'(x)} \to L$ for some $L \in [-\infty,\infty]$, and (ii) Either $f(x) \to 0$ and $g(x) \to 0$ OR $|g(x)| \to \infty$ (no additional condition on $f$)? – Caleb Stanford Aug 17 '16 at 21:44
  • Yes; the only condition on $f$ is that it is differentiable on $(x_0,\infty)$ for some $x_0$. – Mark Viola Aug 17 '16 at 22:20
  • Do you mean the only condition is that $\frac{f'(x)}{g'(x)} \to L$? And this is only in the case that $|g(x)| \to \infty$, right? Is what I said correct? – Caleb Stanford Aug 17 '16 at 22:59
  • Not quite. We need both $f'(x)$ and $g'(x)$ to be differentiable in an open neighborhood of infinity, with $g'(x)\ne 0$ in that neighborhood and $|g'(x)|\to \infty$. Then, if $\frac{f'}{g'}\to L$, then $\frac{f}{g}\to L$. – Mark Viola Aug 18 '16 at 00:07
  • OK, thanks. Well, that is what I said, aside from $g'(x) \ne 0$. That $f', g'$ are differentiable is implicit in the limit statement $\frac{f'}{g'} \to L$. – Caleb Stanford Aug 18 '16 at 00:39
  • You're welcome. My pleasure. -Mark – Mark Viola Aug 18 '16 at 00:43
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    @6005: that $g'(x) \neq 0$ is also implicit in the statement that $f'(x)/g'(x) \to L$. If $g'(x) = 0$ for many large values of $x$ then $f'(x)/g'(x)$ is not defined for many large values of $x$ and hence its limit will not exist. – Paramanand Singh Aug 18 '16 at 08:04
  • @ParamanandSingh thanks for pointing that out! You're right. – Caleb Stanford Aug 18 '16 at 08:09
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The following is not a rigorous answer (good ones were posted already), but rather a note on the intuition behind the result, somewhat related to the last part that you edited into the question.

When the limit exists $\lim_{x\to\infty}f'(x) = L$, intuitively $f(x)$ gets closer and closer to a linear function of slope $L$ as $x\to \infty$ (which can be a horizontal asymptote if $L = 0$ or an oblique one otherwise).

Now, consider a point $P \equiv (x,y)$ on the curve $y = f(x)$. The line through the origin $O \equiv (0,0)$ and point $P$ has the slope $y / x = f(x) / x$. For large enough $x$, the line $OP$ will intersect the asymptote of $f(x)$ in a vicinity of $P$, which becomes smaller and smaller as $x$ grows towards $\infty$ since $f$ "approaches" its asymptote. At the limit $x\to \infty$ the intersection point itself tends to infinity, which is to say that $OP$ tends to a line parallel to the asysmptote i.e. $\lim_{x\to \infty}f(x) / x = L = \lim_{x\to \infty}f'(x)$.

dxiv
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