1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: L'hospital's rule
  1. L'Hospital's Rule (Replies: 3)

  2. L'Hospital's Rule (Replies: 3)

  3. L'Hospital's Rule (Replies: 12)

  4. L'hospitals rule (Replies: 5)

  5. L'Hospital's Rule (Replies: 12)

Loading...