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Quick Limits Question

  1. Dec 22, 2017 #1
    1. The problem statement, all variables and given/known data

    Trying to understand this limit:
    limbio.png
    where ##r>0##
    2. Relevant equations

    I think it's best to proceed by writing this as:

    ## N=1 \pm \frac{\sqrt{Ae^{2rt}}}{\sqrt{1-Ae^{2rt}}} ##



    3. The attempt at a solution

    since ##r>0 ## the exponential term ##\to ## ##\infty## and then since ##A<0## I get two results for ## lim_{t \to \infty} Ae^{2rt} ## depending on ## |A| ##.

    a) If ##|A| < 1 ## it goes to zero.
    if b) ## |A| \geq 1 ## it goes to ##-\infty##

    and where the magnitude of A is not specified in the question.

    If it was however for case a) the limit is of an determinate form: ##1 \pm \frac{0}{1} = 1 ##

    however for b) i get ## 1 \pm \frac{\sqrt{-\infty}}{\sqrt{1+\infty}}## , and I can't see L'Hopitals rule being much use here due to the square root and exponential terms.

    Many thanks in advance.
     
  2. jcsd
  3. Dec 22, 2017 #2

    Ray Vickson

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    $$f(t) = \frac{\sqrt{A e^{2rt}}}{\sqrt{1-Ae^{2rt}}}
    = \frac{e^{-rt}}{e^{-rt}} \frac{\sqrt{A e^{2rt}}}{\sqrt{1-Ae^{2rt}}}
    = \frac{\sqrt{A}}{\sqrt{e^{-2rt} - A} } \to \frac{\sqrt{A}}{\sqrt{-A}}$$
     
  4. Dec 22, 2017 #3
    but A<0, so the numerator is not real?
     
  5. Dec 22, 2017 #4

    Ray Vickson

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    Right, but no less real than the original form ##\sqrt{A e^{2rt}}##.
     
    Last edited: Dec 22, 2017
  6. Dec 22, 2017 #5
    so then do we not get ## 1 \pm i ## rather than ## 1 \pm 1 ## as in the solution above?
     
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