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Limit Problem

  1. Jul 12, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]lim_{x→∞}( \frac{x^x}{(x+1)^x} )^2[/itex]

    2. The attempt at a solution

    Well this limit is actually my most recent step to solving this problem. The initial problem was determining whether or not the following infinite series converges:

    [itex]\sum{(\frac{k}{k+1})^{2k^2}}[/itex]

    The thing I'm curious about is if I can say the following about the above limit:

    As x approaches infinity, [itex]\frac{x^x}{(x+1)^x}[/itex] approaches 1 because the addition of the one becomes less and less significant. Therefore, the entire limit approaches [itex]1^2 = 1[/itex]

    But then again, I was conflicted because the following argument could also be given:

    For all values of x, [itex](x+1)^x[/itex] will be greater than [itex]x^x[/itex]. Furthermore, as x approaches infinity, the difference between [itex](x+1)^x[/itex] and [itex]x^x[/itex] will increase. Therefore, as x approaches infinity, the denominator will grow faster than the numerator, making the limit approach zero.

    Which one of these arguments is correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 12, 2012 #2
    Hi Hertz! :smile:

    That form is of the type 1^(infinity). Do you know how to solve such limit expressions?
     
  4. Jul 12, 2012 #3

    micromass

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    Neither is correct. Are you aware of the following limit?

    [itex]\lim_{n\rightarrow +\infty}{\left(1+\frac{1}{n}\right)^n}=e[/itex]
     
  5. Jul 12, 2012 #4
    Yes I am :)

    I graphed the function and it looks linear (in the positive x) and approaching infinity.. So I'm very interested in seeing how that happened..
     
  6. Jul 12, 2012 #5

    micromass

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    I think you graphed it wrong, then. The function I get is not linear at all!!
     
  7. Jul 12, 2012 #6
    I graphed it with mathematica on the x interval 0 to 10,000. I'll double check, but I'm pretty sure I already double checked when I saw that it was linear haha

    e-
    This is how I imput it:

    (x/(x + 1))^2x

    but apparently this is the correct way:

    (x/(x + 1))^(2x)

    My bad haha
     
  8. Jul 12, 2012 #7
    Thanks for the tip about e, I was able to figure it out with that in mind :)

    Will someone please explain the problems with my original logic though?

    (Here's my solution:)
    [itex]lim_{x→∞}(\frac{x}{x+1})^{2x}[/itex]
    [itex]lim_{x→∞}(\frac{x+1}{x})^{-2x}[/itex]
    [itex]lim_{x→∞}((1+\frac{1}{x})^x)^{-2}[/itex]
    [itex](lim_{x→∞}(1+\frac{1}{x})^x)^{-2}[/itex]
    [itex]e^{-2}[/itex]
    [itex]\frac{1}{e^2}[/itex]
     
  9. Jul 12, 2012 #8

    micromass

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    Let me see if I can make it clear:

    This would be a good approach with limits of things like [itex]\frac{x}{x+1}[/itex]. But one you add an exponent x, then the addition becomes significant again. For example, if x=100, then

    [tex]x+1=101[/tex]

    which is pretty close to x. But

    [tex]x^x=10^{200},~(x+1)^x=2.7*10^{200}[/tex]

    So the adding of the +1 makes the differenc very significant since they are blown up by the exponent x.

    It's not because the denominator grows faster than the numerator, that the limit will become 0. This is exactly an example where this fails.
     
  10. Jul 12, 2012 #9

    Zondrina

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    You're going about this incorrectly. What you're doing without knowing is taking the limit as the SEQUENCE goes to infinity. what you want though is to check what happens when the SERIES goes to infinity which is something completely different.

    Employ the ratio test, if that doesn't work see if the integral test will work.

    Hope that helps.
     
  11. Jul 12, 2012 #10
    Here's what I got when I plotted it - not very linear looking at all and certainly not shooting off to infinity

    Untitled_1.jpg
     
  12. Jul 12, 2012 #11

    Zondrina

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    Does your series start at 0? If so you have an indeterminate term right from the get go and your series diverges. Otherwise if you start anywhere greater than zero your series is going to converge by the ratio test.
     
  13. Jul 12, 2012 #12
    It's just a graph of [itex]\frac{x^x}{(x+1)^x} )^2[/itex] like the OP said he did in post 4
    (if you weren't replying to me then just ignore this o:) )
     
  14. Jul 12, 2012 #13
    As micromass suggested, use this limit. As someone else suggested, use the ratio test to get something that you can massage into something that looks something like a product of these limits.

    EDIT: Woops! I didn't see post number 7 when I posted this. Oh, and I don't know why I called what I did the ratio test, either.
     
    Last edited: Jul 12, 2012
  15. Jul 12, 2012 #14

    SammyS

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    Here's a link to your image:http://postimage.org/image/f52mxsxml/

    Here's the image itself:
    attachment.php?attachmentid=49058&stc=1&d=1342145536.jpg
     

    Attached Files:

  16. Jul 13, 2012 #15
    Thanks for the help guys but the problem was solved clear back in post 7.. I didn't use the ratio test because of the power of k. Instead, I used the root test. That is why I ended up taking this limit in the first place.

    I'm confused, I thought that testing growth rates was one of the great benefits of limits? If you take the ratio of two functions and take the limit, it will approach infinity if the top grows faster, but zero if the bottom grows faster, no?

    Writing this did just spark my memory though, if the limit comes out to a constant then that means the functions have comparable growth rates. Which I guess can be intuitively seen by recognizing that [itex]x^x[/itex] and [itex](x+1)^x[/itex] are the same functions, just one is shifted to the left 1 unit.

    I guess it makes more sense now why the limit would approach a number instead of an extreme like zero or infinity.
     
    Last edited: Jul 13, 2012
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