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Calculating the limits

  1. Dec 7, 2007 #1
    1. The problem statement, all variables and given/known data

    lim[p[tex]\rightarrow[/tex]0] [tex]\frac{ ln(2x^{p}+3y^{p})}{p^2}[/tex]
    1. The problem statement, all variables and given/known data

    I first plugged in zero to see if i can you L'hopital's rule
    I got a (constant)/0 which is not an indeterminant form.

    If I can't use the rule, what should be my next step to find the limit?
    ( I kinda guessed that the the answer's infinity but I'm not sure)
  2. jcsd
  3. Dec 7, 2007 #2
    You don't need L'hopital's rule here since the numerator is always constant... The answer is more straight forward than you think...

    What's lim [x->infinity] 1/(x^2) ?

    edit: didn't notice that x was to the power of p...see office shredder's comment
    Last edited: Dec 7, 2007
  4. Dec 7, 2007 #3


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    The numerator isn't constant, but it approaches one...
    also, the limit is as p goes to 0, not infinity.

    I suspect the easiest way to be rigorous is to bound ln(...) from above and below for small enough p, which gives you that constant/0 term. It's not always infinity though... you haev to be careful of the existence and sign of ln(...) depending on what values x and y take (note they have to be non-negative, or the limit definitely doesn't exist, for example)
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