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Finding limits

  1. Jun 9, 2008 #1
    "x = (e^(2kt) - 1)/(4e^(2kt) - 2)"

    How would I find the limit of this expression as t tends to infinity?

    As t --> infinity, the two exponentials also tend to infinity. However, that was as far as I could go. It is clear by subbing large values of t in, that the limit should be 1/4, however I am unable to prove this.

    Thanks
     
  2. jcsd
  3. Jun 9, 2008 #2

    Vid

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    Factor e^(2kt) out of the top and bottom.
     
  4. Jun 9, 2008 #3

    HallsofIvy

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    One of the first things you should have learned is to divide both numerator and denominator by the "largest" term- in this case e^(2kt) (that's what Vid was saying). You will be left with e^(-2kt) in each and that's easy.

    You could also use L'Hopital's rule here but that is "overkill"
     
  5. Jun 10, 2008 #4

    Dick

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    k is positive, I hope. Otherwise you'd better split into cases.
     
  6. Jun 10, 2008 #5

    Vid

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    If k is negative, the only thing that changes is whether you add or subtract zero; there's no need for cases.
     
  7. Jun 10, 2008 #6

    Dick

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    Add or subtract zero? There are three different cases. k>0, k=0 and k<0.
     
  8. Jun 10, 2008 #7

    Vid

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    My statement about k being positive or negative is true,but I did forget about k=0, which does need its own case.
     
  9. Jun 10, 2008 #8
    Thanks so much for the help. I forget to mention that k is positive. Therefore, I get:

    x = (1-1/e(^2kt)) / ((4 - 2/e(^2kt))

    Thus, as t tends to infinity, x will tend to 1/4.

    However, say k were to be negative, how would it change? Would x tend to 1?

    Thanks
     
  10. Jun 10, 2008 #9

    Dick

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    If k is negative then exp(2kt)->0. Then you can just drop the exponentials and get 1/2.
     
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