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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.

    [tex]\lim_{p\rightarrow\infty}\left(\frac{p^2-p+1}{(p+1)^2}\right)^{2p+3}[/tex]

    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:

    [tex]\ln(L)=\lim_{p\rightarrow\infty}(2p+3)\ln\left(\frac{p^2-p+1}{(p+1)^2}\right)[/tex]

    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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