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Limits problem

  1. May 26, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I tried writing 1-xp as (1-x)(1+x2+x3........xp-1) and same with 1-xq but i don't seem to find any way further.
  2. jcsd
  3. May 26, 2012 #2

    I like Serena

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    Hi Pranav!

    Have you tried turning it into 1 fraction first?

    Since (1-x) is the factor that approaches zero, perhaps you can then divide that away in both nominator and denominator...
  4. May 26, 2012 #3

    Ray Vickson

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    Write [itex]x = 1-r[/itex] and ask for the limit as [itex] r \rightarrow 0.[/itex] Use the binomial expansions of [itex](1-r)^p, \; (1-r)^q[/itex] in the denominators [itex]1-x^p = 1-(1-r)^p \text{ and } 1-x^q = 1-(1-r)^q.[/itex]

  5. May 27, 2012 #4
    Hello ILS!
    At the first start, i did that but i stuck there too so i left it and tried to find a different way.

    Hello RGV!
    Even i try to expand, the denominators become zero again. If i solve 1-(1-r)^p, all the terms will have "r" with them and if i substitute r=0, the denominators again become zero.
  6. May 27, 2012 #5
    You need to combine both ILS and RGV's hints :wink:

    The denominator does give 0, if you put in r=0. But, you can also take out r common from the denominator, and divide it in the numerator, which leaves you a constant term in denominator. Oh, and you need to apply binomial expansion even in the numerator.

  7. May 27, 2012 #6

    I like Serena

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    Well, you kind of have to.
    As it is both fractions approach infinity.
    So you are subtracting infinity from infinity, which is undetermined.

    The only way I now to avoid that, is combining the 2 fractions into 1 fraction first.
    Last edited: May 27, 2012
  8. May 27, 2012 #7

    Ray Vickson

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    Why do you want to put r = 0? You are being asked for a LIMIT as r → 0. That means that you need to find a value, say v, towards which the original expression tends as |r| becomes smaller and smaller and smaller ... . When we compute the limit we never need to put r = 0. In some cases a limit as r → 0 is the same value that would be obtained by actually putting r = 0, but in others (such as the current one in this thread) we cannot ever put r = 0.

  9. May 27, 2012 #8

    Ray Vickson

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    It is much easier to apply the binomial theorem first, to the two separate fractions, take out the common factor r from both denominators, and then combine the fractions. However, either way will lead to the same final answer, provided that one is careful.

  10. May 27, 2012 #9
    Thank you all for the help! :smile:

    Combining the hints provided by ILS and RGV i got my answer.
    Here's the solution:
    (I am dropping the limit word just to make it easier for me to write)

    Substituting x=1-r as r→0,

    pq cancels out. Taking out the common factor from numerator, and substituting r=0, i am left with,

    [tex]\frac{^pC_2q-^qC_2p}{^pC_1 {}^qC_1}[/tex]
    Solving this, i get
    Finally, i get the result
  11. May 28, 2012 #10
    Yep! That's perfect! :smile:
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