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Suppose that $f$ is differentiable for all $x$, and that $\lim_{x\to \infty} f(x)$ exists.

Prove that if $\lim_{x\to \infty} f′(x)$ exists, then $\lim_{x\to \infty} f′(x) = 0$, and also, give an example where $\lim_{x\to \infty} f′(x)$ does not exist.

I'm at a loss as to how to prove the first part, but for the second part, would a function such as $\sin(x)$ satisfy the problem?

user181928
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    $\sin$ is not an example for the second part, since $\lim_{x\to\infty} \sin x$ doesn't exist. – Daniel Fischer Dec 13 '14 at 23:57
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    try something similar, like $\dfrac{\sin(x^2)}x$. From the graph it looks like the slope oscillates between $-1$ and $1$, and it ought to be easy to make this more formal. – Mirko Dec 14 '14 at 00:03
  • @user48481MirkoSwirko and DanielFischer Thank you for your corrections. – angryavian Dec 14 '14 at 00:15
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    For the second part: http://math.stackexchange.com/questions/162078/if-a-function-has-a-finite-limit-at-infinity-does-that-imply-its-derivative-goe – Hans Lundmark Mar 26 '16 at 09:36

2 Answers2

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Assuming both limits exist, apply L'Hospitals rule:

$$L= \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}\frac{e^xf(x)}{e^x}\\=\lim_{x \rightarrow \infty}[f(x) + f'(x)] = L + \lim_{x \rightarrow \infty} f'(x) , $$

and conclude that $ \lim_{x \rightarrow \infty} f'(x)= 0$.

Alternatively, by the MVT there is a point $\xi_x \in(x,x+1)$ such that

$$ f(x+1)-f(x)=f'(\xi_x)$$

If $f'(x) \rightarrow L'$, then for every $\epsilon > 0$ there is a $K>0$ such that $|f'(x) - L'| < \epsilon$ when $x > K$. As $\xi_x > x > K$, it follows that $|f'(\xi_x) - L'| < \epsilon.$

Hence,

$$ L'=\lim_{x \rightarrow \infty}f'(x)=\lim_{x \rightarrow \infty}f'(\xi_x)=\lim_{x \rightarrow \infty}[f(x+1)-f(x)]=0.$$

RRL
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  • I've seen this before, but I'm confused by the $e^x$. Can you please elaborate on the logic behind it? – user181928 Dec 14 '14 at 02:48
  • That's Hardy's trick. However, I added a second proof which is more intuitive. The slope of the secant and the tangent must go to $0$ together. – RRL Dec 14 '14 at 03:12
  • Ohhh well that makes a lot more sense, thanks for clarifying! – user181928 Dec 14 '14 at 03:43
  • what if $\lim_{x\rightarrow \infty}f(x)=0 $? I'm stuck in this case. – ANGIE DUQUE OROZCO Nov 27 '21 at 20:57
  • @ANGIEDUQUEOROZCO: Why is that a problem? Just substitute $0$ for $L$ in either argument. For example, if $0 = \lim_{x \to \infty}f(x) = \lim_{x \to \infty} \frac{e^xf(x)}{e^x}$. We can again apply L'Hopital's rule since the limit of the denominator is $\infty$. Hence $0 = \lim_{x \to \infty} \frac{(e^xf(x))'}{(e^x)'} = \lim_{x \to \infty} \frac{e^xf(x)+e^x f'(x)}{e^x} = \lim_{x \to \infty} (f(x) + f'(x))$. Since the limits of both terms exist we can split as $$0 = \lim_{x\to \infty} f(x) + \lim_{x \to \infty}f'(x) = 0 + \lim_{x\to \infty}f'(x)$$. – RRL Nov 27 '21 at 21:21
  • Even when the numerator is zero? Can I apply the L'Hopital's rule? I have the trouble in that. – ANGIE DUQUE OROZCO Nov 27 '21 at 23:08
  • @ANGIEDUQUEOROZCO: As stated here under General Form we only need the hypothesis that $g(x) \to \pm \infty$ to apply L'Hopital's rule to $\lim \frac{f(x)}{g(x)}$. The limit of the numerator does not even have to exist. In this problem while $f(x) \to 0$ we do not know about the convergence of $e^x f(x)$ unless $f(x)\to 0$ sufficiently fast. Nevertheless, it is irrelevant and we can still apply L'Hopital's rule, since the denominator is $e^x \to \infty$. – RRL Nov 28 '21 at 00:37
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If $\lim_{x\rightarrow\infty}f'(x)=c$ were some positive number, that would imply, for some $0<k<c$ and all large enough $x$ that $f'(x)>k$. Think about what this means intuitively and why this is inconsistent with $f$ converging.

Milo Brandt
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  • So I'm having a bit of trouble with this; would it mean that at some point $f'(x)$ would be less than k? Like I'm thinking of a function whose derivative is, for example, $\frac{2x-1}{x}$. Then choose k, so at some point we will have $f(x) \lt k$? Am I getting there, or am I not on the right track? – user181928 Dec 14 '14 at 01:20
  • We're only interested in when $f'(x)>k$ - which must happen, by definition of a limit, for all $x>N$ for some $N$. So, if the derivative were $\frac{2x-1}x$, we might choose $k=1$ and $N=1$ - meaning that $f'(x)$ was always at least $1$ from there until infinity. – Milo Brandt Dec 14 '14 at 01:30
  • Ok so I found something online: if $f+f′→L$ as $x→∞$ then $f→L$ and $f′→0$ because $lim_{x→∞}f(x) = lim_{x→∞}\frac{e^xf(x)}{e^x} = lim_{x→∞}\frac{e^x(f(x) + f'(x))}{e^x} = lim_{x→∞}(f(x) - f'(x))$ This would tell me that $f'(x)$ has to be zero since we said $f(x)$ is L. I don't quite understand this though; how did $e^x$ come into play? – user181928 Dec 14 '14 at 01:51
  • @user I haven't the faintest idea - that doesn't look like it's effectively proving anything (and much less that it's proving what you're trying to). What I'm trying to get at is if $f'(x)>k$ for all $x>N$, then $f(x)>kx + f(N)$ for $x>N$ - that is, we can bound it below by a linear function. – Milo Brandt Dec 14 '14 at 02:04
  • So I'm effectively going to try and do the same thing to bound it above, then sandwich it to $0$? Sorry if I'm jumping to assumptions, thanks for being very patient with me.. – user181928 Dec 14 '14 at 02:10
  • @user It's not meant as a direct proof - we're trying to show that $f'$ converging to a positive value contradicts the hypothesis that $f$ converges. The sketch of the proof would be, "Suppose $f'(x)$ is not $0$. Then for large enough $x$, the function $f$ is bounded below by a linear function $kx+c$ for positive $k$. This implies $f$ grows arbitrarily large and does not converge. Therefore, if $f$ does converge, $f'$ cannot converge to any positive value" – Milo Brandt Dec 14 '14 at 02:14