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Math Help - If the derivative is bigger than positive constant, is the limit at infinity infinity

  1. #1
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    If the derivative is bigger than positive constant, is the limit at infinity infinity

    Hello

    I'm trying to prove the following: if f is differentiable on all R and there exists a real positive constant c such that f'(x) > c for all x, then the limit of f at infinity is infinity.

    I've managed to prove that if the derivative is positive, then by the Mean Value Theorem the function must be strictly increasing. But I can't seem to manage to prove that the limit isn't some real number...
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  2. #2
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    try this: can you prove that if f'(x) > c then




    The RHS clearly does not converge to an upper limit
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    By the Mean Value Theorem, for every x > 0 we have

    ( f( x ) - f ( 0 ) ) / x = f ' ( t_x ) >= c

    That is,

    f ( x ) >= c x + f ( 0 )

    Now, take limits as x -> + infty .


    Edited: Sorry, I didn't see the previous post.
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  4. #4
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    Thanks to both of you...but I'm not a hundred percent sure why the RHS doesn't converge. Is it because k is a variable and can be as big as we want? If L was a limit of f then f + c would just converge to L + c, right?
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  5. #5
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    Also...might it be enough to say that if f'(x) > c, then f(x) > c*x + k ?
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  6. #6
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    Thanks to both of you...but I'm not a hundred percent sure why the RHS doesn't converge. Is it because k is a variable and can be as big as we want?
    yes

    If L was a limit of f then f + c would just converge to L + c, right?
    Your notation is causing the confusion. F(x) + c converges to L+c, but f(x+c) converges to the same limit as f(x) (i think!).



    Also...might it be enough to say that if f'(x) > c, then f(x) > c*x + k ?
    i dont know how rigorous you're expected to be. It can be shown robustly using the mean value theorum (as demonstrated by Fernando for the case where k=0). of course, it would be easier to use fernandos result directly as he suggested in his post.



    Fernondos solution is better, but i thought this was an interesting alternative approach to the problem:
    Spoiler:

    if f'(x)>c then we can say f'(x)=c + u(x) where u(x) >0

    Now consider f(x1) and f(x2), for arbitrary x1 and x2:



    The integral on the RHS is positive (no need to actually evaluate it!) since u(x) > 0

    Taking the limit as x2 tends to infinity gives the required result for f(x2), and hence f(x). i think.
    Last edited by SpringFan25; April 23rd 2011 at 11:16 AM.
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  7. #7
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    Okay, that looks clever, but I'm afraid I haven't learned much about integrals yet so I'm not quite equipped to understand it...I'll go over it again once I know more Thanks anyway
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