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

  1. #1
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    Limit of the derivative equals zero implies the limit of the function exists

    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:

    $\displaystyle \lim_{x \to \infty} f'(x) = 0 \Longrightarrow \lim_{x \to \infty} f(x) $ exists.
    $\displaystyle \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 $\displaystyle \mathbb{R}\cup\{\pm \infty \} $ In other words, if $\displaystyle \lim_{x \to \infty} \rightarrow \infty$ 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 $\displaystyle f'(x) \rightarrow 0$ as $\displaystyle x \rightarrow \infty$ then f(x) is continuous as $\displaystyle x \rightarrow \infty$
    Thus f(x) has a limit as x $\displaystyle \rightarrow \infty$
    $\displaystyle \qedsymbol$

    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.
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  2. #2
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    Re: Limit of the derivative equals zero implies the limit of the function exists

    Saying that $\displaystyle f'(x)\to 0$ as $\displaystyle x\to \infty$ is not the same as saying that $\displaystyle f$ is differentiable. It is entirely possible that $\displaystyle f$ is not differentiable (and possibly not continuous) for some values of $\displaystyle x$. Suppose that for every open interval $\displaystyle (a,\infty)$, $\displaystyle f$ is not continuous for some $\displaystyle b\in (a,\infty)$. 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 $\displaystyle f$ is differentiable (and therefore continuous) as you stated. Just using definitions, the first statement is:

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

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

    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.
    Last edited by SlipEternal; Feb 16th 2013 at 07:15 PM.
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  3. #3
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    Re: Limit of the derivative equals zero implies the limit of the function exists

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

    Spoiler:
    $\displaystyle f(x)=\sin(\ln(x)), f'(x)=\dfrac{\cos(\ln(x))}{x}$


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

    Spoiler:
    $\displaystyle f(x) = x \sin(\ln(\ln(x))) - \int_e^x {\sin(\ln(\ln(t))) d t}$
    $\displaystyle f'(x) = \dfrac{\cos(\ln(\ln(x)))}{\ln(x)}$
    $\displaystyle \displaystyle \lim_{x\to \infty} \left( \dfrac{x \sin(\ln(\ln(x)))}{x} - \dfrac{1}{x} \int_e^x {\sin(\ln(\ln(t))) d t}\right) = ???$
    I set Mathematica 9.0 to the task of trying to calculate this limit. After almost an hour, it is still calculating.

    Update: After four hours, Mathematica finally gave up. It is quite likely that the limit does not exist, although good luck proving it.
    Last edited by SlipEternal; Feb 17th 2013 at 04:32 PM.
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