Limit of the derivative equals zero implies the limit of the function exists

**jamori** · February 16th 2013, 05:29 PM

I've been tackling a few analysis problems today and have once again hit a bit of a block.

I need to either prove or find a counterexample to the following two propostions:

$\lim_{x \to \infty} f'(x) = 0 \Longrightarrow \lim_{x \to \infty} f(x)$ exists.
$\lim_{x \to \infty} f'(x) = 0 \Longrightarrow \lim_{x \to \infty} \frac{f(x)}{x}$ exists.

Noting here that I am saying a limit exists if it converges in $\mathbb{R}\cup\{\pm \infty \}$ In other words, if $\lim_{x \to \infty} \rightarrow \infty$ then it DOES exist. I know some take an infinite limit as not existing but I am taking it as existing for the purpose of this problem. A non-existent limit would be one like sin(x) where it just sort of oscillates or where the left limit and right limit are not equal.

So, any pointers here? I am pretty sure that both are true because I've had no luck finding a counter example. In a similar problem to this I had earlier (which was effectively the reverse) I was able to employ the Mean Value Theorem.

But basically I am completely stuck.

I thought about saying the following:
We know that if a function is differentiable then it must be continuous. We know that a function is continuous if and only if it has a limit at every point. Hence, if $f'(x) \rightarrow 0$ as $x \rightarrow \infty$ then f(x) is continuous as $x \rightarrow \infty$
Thus f(x) has a limit as x $\rightarrow \infty$
$\qedsymbol$

But I'm pretty sure I'm mis-remembering something with that claim in italics and thus not making any sense at all. Any pointers would be greatly appreciated.

**SlipEternal** · February 16th 2013, 07:05 PM

Saying that $f'(x)\to 0$ as $x\to \infty$ is not the same as saying that $f$ is differentiable. It is entirely possible that $f$ is not differentiable (and possibly not continuous) for some values of $x$ . Suppose that for every open interval $(a,\infty)$ , $f$ is not continuous for some $b\in (a,\infty)$ . Does that negate the possibility of the limit existing for the derivative? I think it implies that there does need to be an open interval about infinity where $f$ is differentiable (and therefore continuous) as you stated. Just using definitions, the first statement is:

$\displaystyle \lim_{x\to \infty} \left( \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \right) = 0 \Longrightarrow \lim_{x\to \infty} f(x) \mbox{ exists}.$

So, that statement is true if and only if there exists $L\in \mathbb{R}\cup\{\pm \infty\}$ such that for all $\varepsilon>0$ , there exists $N\in \mathbb{R}$ such that if $x>N$ then $|f(x)-L|<\varepsilon$ .

Consider what that statement implies. Consider what the derivative implies. Consider contrapositives. Try to figure out exactly what these statements mean before you attempt to prove or disprove anything. I believe if a counterexample exists, it will be to the second statement, and I think the first statement is probably true. A counterexample to the second will probably be a function whose limit is infinity. When you divide by infinity, you may not be able to get any solution.

**SlipEternal** · February 17th 2013, 09:50 AM

Ok, I had my intuition backwards. The first statement has a counterexample. The second probably doesn't.

If you want a derivative to be zero as $x\to \infty$ , you want a bounded function divided by $x$ . So, $\ln(x)$ is unbounded, but its derivative gives $\dfrac{1}{x}$ . Put that into a bounded function. Maybe one that is periodic. Perhaps trigonometric?

Spoiler:

Edit: If the second statement does have a counterexample, this might be it.

Spoiler:

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