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Why limit n-> infinity (3/4)^n = 0

  1. Jul 1, 2013 #1
    Why limit n--> infinity (3/4)^n = 0

    1. The problem statement, all variables and given/known data

    Quick question. Brain flagrance.

    2. Relevant equations

    Why limit n--> infinity (3/4)^n = 0

    3. The attempt at a solution

    ?? Why is this. How do you know.
     
  2. jcsd
  3. Jul 1, 2013 #2

    Mark44

    Staff: Mentor

    Look at the sequence 3/4, (3/4)2, (3/4)3, and so on. What appears to be happening?
     
  4. Jul 1, 2013 #3
    Got it. What about this? THanks
    limit (1+k/n)^n as n->infinity

    Just go here http://www.calcchat.com/book/Calculus-ETF-5e/
    You have to put in chapter 9 section 1 and question 69. I just don't understand how they knew to make a substitution.
     
  5. Jul 1, 2013 #4
    Ps then after they make the sub with u they write it as u goes to 0....huh?
     
  6. Jul 1, 2013 #5

    Mark44

    Staff: Mentor

    They are likely using a result from a previous problem. If so, the purpose of the substitution was to get the problem in that form.
     
  7. Jul 1, 2013 #6

    Mark44

    Staff: Mentor

    For the substitution u = k/n, if n → ∞, what does u do?
     
    Last edited: Jul 2, 2013
  8. Jul 1, 2013 #7
    Stays the same? Isn't k a constant? So OH lol so you take the limit as u goes to 0 because as k/n goes to infinity u goes to 0? I say that right?
     
  9. Jul 2, 2013 #8

    Mark44

    Staff: Mentor

    Sorry, I was typing faster than I was thinking. I meant "what does u do" as n → ∞?(I edited my earlier post.)
    No, you didn't. As k/n goes to infinity, so does u, since u = k/n. The way to look at it is, as n → ∞, then the fraction k/n → 0.
     
  10. Jul 3, 2013 #9
    This always bugs me, because assuming k > 0, as n → ∞, k/n → 0 from the right.
     
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