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Limits: showing your working.

  1. Jun 11, 2008 #1
    eg. y = 5ke^(-0.2t) - 5k + 2

    How would one show the working to show what happens as t tends to infinity? Would something like this be ok?

    as t--> infinity

    e^(-0.2t) --> 0

    therefore 5ke^(-0.2t) ---> 0

    therefore 5ke^(-0.2t) - 5k + 2 --> 5k + 2

    therefore y ---> 5k + 2

    Thanks
     
  2. jcsd
  3. Jun 11, 2008 #2

    Defennder

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    Looks ok. But you left out a minus sign.
     
  4. Jun 11, 2008 #3

    tiny-tim

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    Hi nokia8650! :smile:

    I guess it means a delta-epsilon proof. :smile:
     
  5. Jun 12, 2008 #4

    HallsofIvy

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    No, I would not say that "show the working" means a "delta- epsilon" proof. What You have done is correct. You might add something like "Since f(x)= ex increases without bound as x increases, e^(-0.2t) --> 0 as t --> infinity".
     
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