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Limit of (3/4)^(n+1) as n approaches infinity

  1. May 11, 2015 #1
    Hello,

    I was just wondering how to solve this limit: Limit of (3/4)^(n+1) as n approaches infinity

    My attempt:
    (3/4)^(n+1) = (3^ (n+1) ) / (4 ^ (n+1))
    Top goes to infinity and bottom goes to infinity to use l'hopital rule.
    lim = ( ln(3) * 1 * 3^(n+1) ) / ( ln(4) * 1 * 4^(n+1) )

    But the correct answer is apparently 0.
    Not sure if i'm totally wrong or if I'm just halfway there.

    Thanks
     
  2. jcsd
  3. May 11, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You are not done here. Both numerator and denominator still go to infinity, and you didn't gain anything.

    There are more direct approaches that work.
     
  4. May 11, 2015 #3
    Am I overthinking this.
    Is it as simple as the denominator is larger and gets bigger faster and numerator making the whole thing = 0 for that reason?
     
  5. May 11, 2015 #4

    Mark44

    Staff: Mentor

    Yes.
    Something like that, but the "whole thing" never equals zero. It gets arbitrarily close to zero, though.
     
  6. May 11, 2015 #5
    OK I'll use that logic then. thx
     
  7. May 11, 2015 #6

    Mark44

    Staff: Mentor

    If a is any number in the interval (-1, 1), then ##\lim_{n \to \infty}a^n = 0##.

    BTW, please don't delete the three parts of the homework template. They are there for a reason; namely, to help you organize your problem and the work you've done.
     
  8. May 11, 2015 #7
    I see thanks again.
     
  9. May 11, 2015 #8
    if you write (3/4)^(n+1) as ##e^{(n+1)\ln{(\frac{3}{4})}}## maybe it will be easier to see why it goes to zero since the ln is negative & you don't need l'hopital's rule either.
     
    Last edited: May 11, 2015
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