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Homework Help: Is the limit also infinity?

  1. Feb 4, 2006 #1
    Hi,



    What is the limit of
    [tex]ln(\frac{1+1/ \sqrt(1+1/R)}{1-1/ \sqrt(1+1/R}) [/tex]

    as R reaches infinity.


    (latex did not show it very well. But the numerator is same as denominator except there is a + sign instead of a - sign.

    As R reaches infinity the argument of the ln reaches infinity. Is the limit also infinity?





    Thank You.

    Gamma.
     
    Last edited by a moderator: Feb 5, 2006
  2. jcsd
  3. Feb 5, 2006 #2

    VietDao29

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    Homework Helper

    If your LaTeX image is big, don't use [ itex ], use [ tex ] instead.
    So do you mean:
    [tex]\lim_{R \rightarrow \infty} \ln \left( \frac{1 + \frac{1}{\sqrt{1 + R}}}{1 - \frac{1}{\sqrt{1 + R}}} \right)[/tex]? Or what?
    If you mean that, then if [tex]R \rightarrow \infty[/tex], then [tex]\sqrt{1 + R} \rightarrow \ ?[/tex], [tex]\frac{1}{\sqrt{1 + R}} \rightarrow \ ?[/tex], [tex]1 \pm \frac{1}{\sqrt{1 + R}} \rightarrow \ ?[/tex].
    Can you go from here? :)
     
  4. Feb 5, 2006 #3

    HallsofIvy

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    Gamma, I've taken the liberty of editing your Latex by changing "itex" to "tex". "itex" doesn't work well with complex fractions.

    As VietDao29 told you- the argument inside the ln does NOT go to infinity. It should be sufficient to see what happens to 1/R as R goes to infinity.
     
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