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Study the continuity of this function

  1. Dec 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Study the continuity of the function defined by:
    ## \lim n \to \infty \frac{n^x-n^{-x}}{n^x+n^{-x}}##

    3. The attempt at a solution

    I've never seen a limit like this before.
    The only thing I have thought of is inserting random values of x to see it the limit exists.
    For instance, in this case, for x=0 I'd have ##\frac{\infty^0-\infty^0}{\infty^0+\infty^0}##

    which means the function doesn't exist. but every other value of x, it's okay.
    Or am i supposed to solve the limit? (btw, how can I solve a limit for n to infinity??)
    thank you
     
  2. jcsd
  3. Dec 27, 2012 #2

    jbunniii

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    By definition, [itex]n^0 = 1[/itex] for any nonzero [itex]n[/itex], so your fraction reduces to
    [tex]\frac{n^0 - n^0}{n^0 + n^0} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0[/tex]
    What does this imly about the limit
    [tex]\lim_{n \rightarrow \infty} \frac{n^0 - n^0}{n^0 + n^0}[/tex]?
     
  4. Dec 28, 2012 #3
    well, I'd say it means the limit exists and it is = 0 (because I can evaluate it before adding the infinities in the equation).
    so the function should continuos on all ##\mathbb{R}##.
    Or actually, how should i work for x→∞?
    sorry if i have so many doubts about a simple question, but i have never seen this kind of limits before
     
  5. Dec 28, 2012 #4

    Dick

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    It's not that hard. You just have to think through all the cases. Suppose x=1. Try and figure it out without just substituting 'infinity' for n. That's not very informative. Just put x=1.
     
    Last edited: Dec 28, 2012
  6. Dec 29, 2012 #5
    ok.
    I see that whatever x i choose (except x=0), I always get a 0 in the nominator and an infinity in the denominator, so appartently f(x)=0. But I'm not sure about that..
     
  7. Dec 29, 2012 #6

    jbunniii

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    Let's take a simple concrete case, say x = 1. Then
    [tex]\frac{n^x - n^{-x}}{n^x + n^{-x}} = \frac{n - 1/n}{n + 1/n} = \frac{1 - 1/n^2}{1 + 1/n^2}[/tex]
    What is the limit of this expression as [itex]n \rightarrow \infty[/itex]?
     
  8. Dec 30, 2012 #7
    for x=1 the limit is 1.
    and for x=2 as well, because
    ##\frac{n^2-n^{-2}}{n^2+n^{-2}}##dividing both members by## \frac{1}{n^2} =\frac{1-\frac{1}{n^4}}{1+\frac{1}{n^4}}## and so forth ##\forall x \in \mathbb{R}## except x=0, where f(x)=0.
    Then, can i say it is continuos on all |R except between 0 and 1?
     
  9. Dec 30, 2012 #8

    Dick

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    You'd better try some more x values before you state a conclusion. What about x=(1/2) or x=(-1)?
     
  10. Dec 30, 2012 #9
    Ok.
    I've tried several values and found out that if
    ##x>0 \Rightarrow f(x)=1##
    ##x=0 \Rightarrow f(x)=0##
    ##x<0 \Rightarrow \lim f(x)= \frac{\infty}{\infty}## so the function doesn't exist there
    Also, it is discontinuos in 0.

    Are my assumptions right?
     
  11. Dec 30, 2012 #10

    Dick

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    Can you show how you reached that conclusion for x<0? Try x=(-1). infinity/infinity doesn't necessarily mean the limit is undefined.
     
  12. Dec 30, 2012 #11
    for x=-1 I get
    ## \frac{1-\frac{1}{n{^-2}}}{1+ \frac{1}{n^{-2}}}##
    ##\frac{1-n^2}{1+n^2}= \frac{\infty}{\infty}##
    If ##\frac{\infty}{\infty}## doesn't mean it's undefined, when can i say the function is discontinuos?
     
  13. Dec 30, 2012 #12

    Dick

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    Divide numerator and denominator by n^2.
     
  14. Dec 30, 2012 #13
    right, i didn't notice it, then for x<0 i get f(x)=-1.
    what can i say if i get ##\frac{\infty}{\infty}##? and when can i say that the function is not continuos, if that's not enough?
     
  15. Dec 30, 2012 #14

    Dick

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    You can't say anything just from looking at ##\frac{\infty}{\infty}##. A form like that might have a limit and it might not. Ok so f(x)=(-1) for x<0, f(0)=0 and f(x)=1 for x>0. What about continuity?
     
  16. Dec 30, 2012 #15
    well, as the right limit is different from the left one, I'd say 0 is a discontinuity point (the only one in R).
    fortunately, there are no infinity/infinity cases in this function. but if one of these cases happens with a similar limit, should i just leave it? (if algebric manipulation can't help)
    thank you very much for your help, anyway :)
     
  17. Dec 30, 2012 #16

    Dick

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    Yes, it's discontinuous only at 0. Leaving a limit as ∞/∞ is about the same thing as saying 'I don't know what the limit is'. If algebra doesn't help you resolve it then there are other tools like l'Hopital.
     
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