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B Can f(x) and f'(x) both approach a non-zero constant?

  1. Mar 26, 2017 #1
    Hello everyone! I've been curious about this for a while and couldn't come to a conclusion on my own so I've decided to ask it here.

    I'm wondering if it's possible for a function, f(x), to have a rule which would allow it and it's derivative to both approach a constant value as x approaches infinity. For example, ##lim_{(x→∞)} f(x) = 3##, ##lim_{(x→∞)} f'(x) = 5##. Note, I don't mind if the constants are equal to each other, the only thing that's important is that they are non-zero constants.

    Thank you all for your time :)
     
  2. jcsd
  3. Mar 26, 2017 #2
    Not that I can think of. The first derivative is a slope, implying that the function changes and would not approach a constant value. If f(x) were to approach a constant value, the slope of the function would be approaching zero for that domain.
     
  4. Mar 26, 2017 #3
    Thank you for your advice. I had figured as much, but I still felt it necessary to ask in case there is a function that would act as an exception to the general rules that I don't yet know about.
     
  5. Mar 26, 2017 #4
    You can use inequalities to prove the impossibility. If

    [tex]
    \lim_{x\to\infty} f'(x) = 5
    [/tex]

    then there exists [itex]R[/itex] such that [itex]f'(x)>4[/itex] for all [itex]x>R[/itex]. Then also

    [tex]
    f(x) > f(R)+ 4(x-R)
    [/tex]

    holds for all [itex]x>R[/itex]
     
  6. Mar 27, 2017 #5
    How about something like 3+5*(x mod 1) ?
    Say 3+5*(x mod (1/x)) ?
    The limit is 3 and f'(x)=5, most of the time anyway.
     
  7. Mar 27, 2017 #6
    That's very close to what I'm looking for! But not quite. As ##x→∞##, ##f(x)→8##. However ##f'(x)## only approaches ##5## as ##x→-∞##. Is there a further step to manipulate this so that they both approach a constant as ##x## approaches positive infinity?
     
  8. Mar 27, 2017 #7
    Maybe your definition of "mod" is different from mine?
    I'm pretty sure that f'(x) is exactly 5 everywhere, and also ##3\leq f(x) \leq 3+5/x## which approaches 3 as x goes to positive infinity.
     
  9. Mar 27, 2017 #8
    It is not clear what you mean with that mod function. Can you write the same things in terms of the floor function [itex]x\mapsto \lfloor x\rfloor[/itex]?

    Anyway, it looks like you are speaking about functions that are not differentiable at all points. Why would you be interested in the limit

    [tex]
    \lim_{x\to\infty} f'(x)
    [/tex]

    if [itex]f[/itex] was not differentiable at least in some set [itex]]R,\infty[[/itex] with some [itex]R\in\mathbb{R}[/itex]?
     
  10. Mar 28, 2017 #9
    $$f(x)=3+5(x^2-\lfloor x^2\rfloor)/x$$
    It was the only way I could figure out to satisfy the requirement of the original poster.
     
  11. Mar 28, 2017 #10
    It certainly is the only way to get a some kind of function requested by the original poster, since I already proved (or showed some relevant steps of a proof) that such function cannot exist if it is required to be differentiable for all sufficiently large [itex]x[/itex]. The inequality in my proof can be justified by the mean value theorem, by the way, I forgot to mention it above.
     
  12. Mar 28, 2017 #11
    I guess the original question has already been answered, but anyway, I could not help taking a look at the function you just defined, and I guess of course you have to keep doing some exercises to maintain your math skills :-p

    The formula

    [tex]
    x\mapsto \frac{x^2 - \lfloor x^2\rfloor}{x}
    [/tex]

    defines a function that has discontinuities at points [itex]x=\sqrt{1},\sqrt{2},\sqrt{3},\ldots[/itex], and is differentiable between these points. When [itex]x[/itex] assumes values from an interval [itex]\sqrt{n}\leq x < \sqrt{n+1}[/itex], the expression assumes values from the interval

    [tex]
    0\leq \frac{x^2 - \lfloor x^2\rfloor}{x} < \frac{1}{\sqrt{n+1}}
    [/tex]

    So these values are going to zero in the limit [itex]n\to\infty[/itex].

    The derivative is

    [tex]
    D_x \frac{x^2 - \lfloor x^2\rfloor}{x} = 1 + \frac{\lfloor x^2\rfloor}{x^2}
    [/tex]

    When [itex]x[/itex] assumes values from an interval [itex]\sqrt{n}< x<\sqrt{n+1}[/itex], the derivative assumes values from the interval

    [tex]
    1 + \frac{n}{n+1} < 1 + \frac{\lfloor x^2\rfloor}{x^2} < 2
    [/tex]

    The lower bound can be written in the form

    [tex]
    2 - \frac{1}{n} + O\Big(\frac{1}{n^2}\Big)
    [/tex]

    so we see that the values of the derivative get squeezed close to 2 at the limit [itex]n\to\infty[/itex].
     
  13. Mar 28, 2017 #12

    PeroK

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    Another approach is to have a function that is not defined everywhere. E.g a function that is defined on a sequence of diminishing open intervals.

    In this case you could have a function that is continuous and differentiable everywhere ( on its domain) but meets the criteria.
     
  14. Mar 28, 2017 #13
    Are you sure about this? I might have made a mistake in rewriting x mod (1/x) into the floor function but I'm pretty sure that
    $$\frac{d}{dx} (x\ \mod\ \text{anything}) = \frac{d}{dx} (x) = 1$$
    except on points where the mod function does its thing.

    Edit: Well you're right, it looks like something funny is going on here when there's a function of x in the denominator.

    http://www.wolframalpha.com/input/?...mod+2,+x+mod+(1/2),+x+mod+(6/x)}+from+6+to+16
     
    Last edited: Mar 28, 2017
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