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Exponential of this limit

  1. May 25, 2007 #1
    1. The problem statement, all variables and given/known data
    evaluate :
    [tex] \lim_{x\longrightarrow1}\ \left\ ( \frac{p}{1-x^p} - \frac{q}{1-x^q} ) \ \right p,q\in N [/tex]

    2. What I've tried.

    I s'pose L`Hopital's Cant be applied as this is not 0/0 form.
    I took the LCM and put them in this form :

    [tex] \lim_{x \longrightarrow 1}\ \frac{p-px^q -q+qx^p}{1-x^p-x^q+x^{p+q}} [/tex]

    But still i cant replace x-1 by h. How do i proceed
    Thx
     
    Last edited: May 25, 2007
  2. jcsd
  3. May 25, 2007 #2
    take the exponential of this limit, and then youll have a limit with which you can calculate it via lhopital's rule.
     
  4. May 25, 2007 #3
    Sorry i dont understand what taking exponential means. Could you plz explain ?
    Thx
     
  5. May 25, 2007 #4
    well, calculate the limit of e^([p/(1-x^p)]-[q/(1-x^q)]), and use the fact that: lim e^g(x)=e^(lim (g(x))). (this is correct only if the limit is finite).
     
  6. May 25, 2007 #5
    THx for the help, but how do i simplify the exponent? And since the denominator is constant, how do i apply lhopitals rule?
     
  7. May 25, 2007 #6

    Dick

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    Try another approach. Since x->1, write x=1+y where y->0. Now write x^p as (1+y)^p and use the binomial expansion. Since y->0 you can ignore a lot of the higher powers in y. In this case you'll only need to keep the first three terms.
     
  8. May 25, 2007 #7
    Thx, that worked perfectly. I simplified and got (p-q)/2 which is the right answer.
    But i have one question to ask, how did you decide to restrict to only 3 terms of the binomial expansion?

    Once again, thx a lot
     
  9. May 25, 2007 #8

    Dick

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    Because I tried keeping just one term and it cancelled exactly. Then realized I'd better keep another term. You can keep all of the terms if you want. But then you'll realize that the contributions beyond the third term go to zero as y->0.
     
  10. May 25, 2007 #9
    Yeah, fine.
    Thank you very much
     
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