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Limit question

  1. Feb 27, 2009 #1
    [tex]
    \lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
    [/tex]

    i get infinity - infinity
    its undefined
    ??
     
  2. jcsd
  3. Feb 27, 2009 #2

    tiny-tim

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    Hi transgalactic! :smile:

    It's easier to read if you write …

    [tex]\lim _{x->+\infty} (x+2)^{2/3} -(x-2)^{2/3}[/tex] :wink:
    eek! you're not allowed to do that! :eek:

    Hint: write it [tex]\lim _{x->+\infty} x^{2/3}\left((1 + 2/x)^{2/3} -(1 - 2/x)^{2/3}\right)[/tex] :wink:
     
  4. Feb 27, 2009 #3

    Mark44

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    infinity - infinity is one of several indeterminate forms. None of them are ever an answer.
     
  5. Feb 27, 2009 #4

    djeitnstine

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    A hint when doing limits, just think about what happens when the number gets REALLY big. Eventually does it matter that you add 2 or subtract 2 from say...10000000000000000? It will approximate to [tex]x^{\frac{2}{3}}-x^{\frac{2}{3}}[/tex] in the end...

    Also the result is the same as tinytim's =]
     
  6. Feb 27, 2009 #5
    so i get 0*infinity

    its not defined either
     
  7. Feb 27, 2009 #6

    tiny-tim

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    Nooo! :cry: … expand the bracket first, then multiply by the x2/3, then take the limit.
     
  8. Feb 27, 2009 #7
    (x-2)2/3 is less than (x+2)2/3 so each time the x gets larger a larger sum is in (x-2) than(x+2) so it goes down... to minus infinity.
     
  9. Feb 27, 2009 #8

    djeitnstine

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    Noo that is also not the result! Read my post and tinytim's again

    When in doubt graph it!
     
  10. Feb 27, 2009 #9

    HallsofIvy

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    When tiny-tim says "expand the brackets first" I believe he is referring to the generalized binomial expansion. And you should only need the first few terms.
     
  11. Feb 27, 2009 #10

    Mark44

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    Another approach is to multiply the original expression by 1 like so:
    [tex](x + 2)^{2/3} - (x - 2)^{2/3} \frac{(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}} {(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}}[/tex]

    This looks pretty complicated, but it's based on a relatively simple idea: that (a - b)(a2 + ab + b2) = a3 - b3.

    After using this technique, the numerator can be simplified to a single term, and the limit of the whole expression can be found without much trouble.
     
  12. Feb 27, 2009 #11

    djeitnstine

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    Why need to get so complicated :S this is quite a simple problem. Quite frankly the answer is 0. There is an asymptote in the graph tending towards 0 as x increases towards positive infinity. Now which result out of mine or tinytim's did not explain this?

    I can re iterate of course "Eventually does it matter that you add 2 or subtract 2 from say...10000000000000000? It will approximate to [tex] x^{\frac{2}{3}}-x^{\frac{2}{3}}[/tex] in the end..." which of course is always tending towards 0...
    124h3pt.jpg
     
  13. Feb 28, 2009 #12

    Mark44

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    Agreed that this is a simple problem, and that the limit is 0. Tiny Tim's approach is superior to mine in its simplicity. Your approach, using a computer-generated graph, is useful for gaining understanding about the behavior of this function, but that's about it. It is well-known that computers don't produce exact results, because of the limited precision they use in their calculations, and especially so in raising to fractional powers. And in some cases, they can produce results that are incorrect (e.g., the earlier Intel Pentium math coprocessors back in 1992 or so). You also said that for large x, x + 2 was about the same as x - 2. This is correct, and is helpful in understanding, but it should be said in mathematical terms. Otherwise, it's akin to arm-waving, IMO.

    My approach is different from, and slightly more complicated, than Tiny Tim's. But so what? Multiplying by a factor to produce a difference of cubes isn't all that complicated, and the intent was to show another method of approach that might be useful in other problems.

    The more tools you have to work with, the more problems you're equipped to solve.
     
  14. Feb 28, 2009 #13

    djeitnstine

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    Well I do understand that your method was valid, however I believe when helping someone whom is slightly confused the simplest approached should be used. Don't you agree?
     
  15. Feb 28, 2009 #14

    Mark44

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    I agree that the simplest approach should be shown first. Then you can show other approaches, just as was done in this thread.
     
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