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Homework Help: Limit log rule help

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Limit as p->infinity of ((p^2-p+1)/((P+1)^2)^(2p+3)

    In case the parenthesis are confusing, its one giant fraction all raised to the (2p+3) power.

    2. The attempt at a solution
    I set the entire problem equal to L and took the ln of both sides. This lets me move the power down using a log rule So:

    ln(L) = lim p->infinity (2p+3)* ln((p^2-p+1)/((P+1)^2)

    Using the log rule of ln(m/n) = ln(m) – ln(n):

    lim p-> infinity (2p+3)[ln(p^2-p+1) - 2ln(p+1)]

    At this point Im not sure. I think I can put it into a form where I can then use hopital's rule? So:

    [ln(p^2-p+1) - 2ln(p+1)] / (1/(2p+3))

    Will that help? I took the derivative of top and bottom but its not looking like something I can use.
  2. jcsd
  3. Feb 25, 2009 #2

    Tom Mattson

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    Re: limit

    Let's clean this up a bit using LaTeX. If you want to see the code that I used to generate this image, just click on it. Make sure popups are allowed though.


    So far so good.

    This isn't going to help too much because you get the indeterminate form [itex]\infty - \infty[/itex]. Instead write it like this:


    This is the indeterminate form [itex]\infty \cdot 0[/itex]. Do you know how to handle that?
  4. Feb 25, 2009 #3
    Re: limit

    That latex code confuses me, but hopefully I can still write it out like ya would into a calculator.

    So, with 0 * infinity I need to make it look like this:

    ln((p^2-p+1)/((P+1)^2) / (1/(2p+3))

    Ordinarily I would use hopital's rule, but wouldnt the top fraction become uglier after differentiating rather than more helpful?
  5. Feb 26, 2009 #4

    Tom Mattson

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    Re: limit

    It gets uglier before it gets better. If you take the derivative of the top and the bottom you end up with a ratio of rational expressions, which is itself a rational expression. Limits of rational expressions at infinity are easy to compute.
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