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Lim (2/3)^n x-> infintiy

  1. Mar 13, 2005 #1
    1) lim (2/3)^n
    x-> infintiy
    2) lim (4/3)^n
    x-> infinity
     
  2. jcsd
  3. Mar 13, 2005 #2
    1) the answer is 0
    because 2/3 = 0.6666 < 1 ---> lim=0

    2) the answer is 1
    because 4/3 = 1.3333
    and when this number tends to infinity it is considered equal to one thus the answer is 1
     
  4. Mar 13, 2005 #3
    i forgot to tell you also that u made a mistake n should tend to infinity and not x. :)
    joe
     
  5. Mar 13, 2005 #4

    dextercioby

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    The first limit is [tex] +\infty [/tex]...

    Daniel.
     
  6. Mar 13, 2005 #5

    chroot

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    Uhhh... what? :uhh:

    Absolutely false. The answer to this limit is infinity.

    If you multiply a quantity by a fraction less than one an infinite number of times, it tends to zero.

    If you multiply a quantity by a number greater than one an infinite number of times, it tends to infinity.

    In no way, whatsoever, is 4/3 ever "considered equal to one."

    - Warren
     
  7. Mar 13, 2005 #6
    [tex]\lim_{x\rightarrow\infty} (\frac{4}{3})^x = \infty[/tex]

    If you can't see that, graph the equation. Or if you don't have a calulator, use the first derivative for critical points, and you find none.

    The first answer is correct, but the second one is infinity.


    Jameson
     
  8. Mar 13, 2005 #7
    yeah, i doubted it first.. i thought it was wrong. You corrected it.
    thanks guys :)
     
  9. Mar 13, 2005 #8

    dextercioby

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    It can be elegantly proven using the famous commutation rule:

    [tex] \ln \lim_{x}f(x) =\lim_{x}\ln f(x) [/tex]

    Daniel.
     
  10. Mar 14, 2005 #9
    No its not,
    1) lim (2/3)^n
    x-> infintiy
    =2^n/3^n

    so its zero.
     
  11. Mar 14, 2005 #10

    dextercioby

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    I was referring to post #2...:tongue:

    Daniel.
     
  12. Mar 14, 2005 #11
    That's odd, I have never heard of this (maybe I did, but have forgotten). Time to dig up the ol' calculus book.
     
  13. Mar 14, 2005 #12

    dextercioby

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    It comes about because the natural logarithm is an strictly ascending function on its domain.


    Daniel.
     
  14. Mar 14, 2005 #13

    Hurkyl

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    You're forgetting the most important part -- continuity. :tongue2:
     
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