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Finding the limit of a velocity vector

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle moves in the XY plane so that at any time t is greater than or equal to 0 its position (x,y) is given by x = e^t + e^-t and y = e^t - e^-t.

    Find the limit as t approaches infinity of dy/dx.

    2. Relevant equations

    I found dy over dx to be e^t + e^-t over e^t - e^-t

    3. The attempt at a solution

    I plugged in infinity for t but it comes out as infinity over infinity, and then I tried L'Hôpital's rule and then multiplied by the reciprocal, but still nothing. I dont know where to go from here.
  2. jcsd
  3. Feb 25, 2010 #2
    As t goes to infinity, e^(-t) goes to zero, so you can neglect those terms.
  4. Feb 25, 2010 #3
    Yeah I figured that, but that leaves me with e^t over e^t which ends up as infinity
  5. Feb 25, 2010 #4
    The limit of e^t/e^t is just 1. Yes, "infinity over infinity" is an indeterminate form, but that just means that, for arbitrary limits that both go to infinity, you can't tell what the limit of their quotient is. If you know what the limits are, then you often can get an answer.

    Look, e^t/e^t=1 for any finite t. How could it have any other limit than 1 at infinity (or anywhere else)? To make it rigorous, you could put a bound on the difference between your original quotient and this one, and show that the difference goes to 0 as t goes to infinity.
  6. Feb 25, 2010 #5
    oh yeah. hahahahaha whoops. thanks man
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