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Homework Help: L'hospital's rule

  1. Feb 1, 2007 #1
    I have the equation A
    44*e^(kt/5)+49((1-e^(kt/5))/k)

    and I'm supposed to evaluate as k-->0

    I think I'm supposed to apply l'hospital's rule to the second part of the equation, which would give
    49*((1-t/5*e^(kt/5))/1)
    which as k-->0 is
    49*(1-t/5)

    so the whole thing as k-->0 is
    44+49*(1-t/5)

    This isn't right, and I also tried l'hosital's rule on the first part of A, which would give 44*t/5 and this isn't right either.

    What am I doing wrong?

    Thanks.

    Here's the whole question, in case I'm not reading it right:
    Find the limit of this velocity for a fixed time t_0 as the air resistance coefficient k goes to 0.
     
  2. jcsd
  3. Feb 1, 2007 #2

    arildno

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    You have, by L'Hopital's rule:
    [tex]\lim_{k\to{0}}49\frac{1-e^{\frac{kt}{5}}}{k}=\lim_{k\to{0}}49\frac{-\frac{t}{5}e^{\frac{kt}{5}}}{1}=-\frac{49}{5}t[/tex]
     
  4. Feb 1, 2007 #3
    Oh yeah, that 1 should've disappeared from what I found.

    Thanks a lot.
     
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