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Question on limit at infinity

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Hello everyone, I am just new to this forum and also a beginner at calculus.
    I have a question from my textbook. It's:
    Find an example of f(x) that satisfies the following conditions :
    f(x) is differentiable for all x>0;
    limx->∞f(x) =2;
    limx->∞f'(x) does not exist;

    I think that if f(x) satisfies the second condition it must have a horizontal tangent at infinity,which means f'(x) = 0 at infinity, am I right? and what does "f'(x) does not exist" really mean?
    Thanks in advance.
    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 26, 2011 #2

    LCKurtz

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    You are right about the horizontal asymptote (not horizontal "tangent") of y = 2 as x → ∞. And many graphs you have seen to have the curve "leveling out" as the graph approaches the asymptote, in which case you would have

    [tex]\lim_{x\rightarrow \infty}f'(x)=0[/tex]

    What you need to do is find an example that has the y = 2 asymptote but the slope doesn't get close to 0, maybe because it "wobbles back and forth", to phrase it informally.
     
    Last edited: Sep 26, 2011
  4. Sep 26, 2011 #3

    SammyS

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    Welcome to the PH Forums !!!

    It doesn't say f'(x) does not exist.

    It says limx➙∞f'(x) does not exist.

    If you're just beginning Calculus, this problem could be difficult.

    You are correct as far as rational functions are concerned.

    Try to modify a sine or cosine function, so that its amplitude decreases as x➙∞ , but oscillates more and more rapidly as x➙∞ . If it oscillates rapidly enough, then the derivative may oscillate with constant or increasing amplitude. Therefore, the limit derivative will not converge as x➙∞ .

    See the image of a graph of such a first derivative, f'(x). attachment.php?attachmentid=39296&stc=1&d=1317059429.gif ...Plotted in WolframAlpha.
     

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  5. Sep 26, 2011 #4
    Actually I confused "tangent" with "asymptote"... Thanks for the help from both of you.
     
  6. Sep 26, 2011 #5

    SammyS

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    So, what did you come up with for f(x) ?
     
  7. Sep 27, 2011 #6
    I came up with this :
    f(x)=2+sin(x3)/x
    so f'(x) = sin(x3)/x2-3cos(x3) and it is not stable at infinity because of cos(x3),right?
    Please tell me if it is ok or not. Thank you.
     
  8. Sep 27, 2011 #7

    SammyS

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    Last edited by a moderator: Apr 26, 2017
  9. Sep 27, 2011 #8
    Thank you.I should have done more exercises.
    and also thanks for introducing WolframAlpha to me. That's really helpful.
     
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