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Determine the limit of the convergent sequence

  1. Jul 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the limit of the convergent sequence:
    ##a_n## =##"3/n" ^ "1/n"##
    http://www.wolframalpha.com/input/?i=lim+as+x+approaches+infinity++%283%2Fx%29^%281%2Fx%29


    2. Relevant equations

    --

    3. The attempt at a solution[/b

    So I tried to get this series to look like 0/0 or ∞/∞ so I could use the L'Hopital rule.
    I took the Ln of both sides. I got (ln(3/n))/n. As n approaches ∞, the numerator is -∞ and the denominator is ∞. Can
    I use the L'Hopital's rule now?


    THANKS!
     
    Last edited: Jul 4, 2012
  2. jcsd
  3. Jul 4, 2012 #2

    Zondrina

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    Given the way you've presented the sequence, this rule will be extremely useful to you :

    (a/b)^c = a^c / b^c

    Big hint : Remember that 1 over something really really big... aka ∞ tends to 0.

    Example :

    lim x->∞ 1/x = 0
     
  4. Jul 4, 2012 #3
    Thank you,
    So I get (3^(1/n)) / (n^(1/n))
    If n approaches ∞, the the numerator is 1 because the exponent approaches 0.
    So would it be sufficient for me to apply the same mentality to denominator and say
    that it approaches 1? That would give me 1 for the limit of the sequence which is the correct
    answer.
     
  5. Jul 4, 2012 #4

    Zondrina

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    Yup :) good job. Sometimes the most basic rules can save you the most headaches.
     
  6. Jul 4, 2012 #5

    I like Serena

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    Welcome to PF, ArthurRead! :smile:

    Yes...
     
  7. Jul 4, 2012 #6

    I like Serena

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    And no, you cannot just say that ##n^{1/n}## approaches 1.

    Consider that ##n^{1/2}## approaches infinity.
    And that ##2^{1/n}## approaches 1.
    ##n^{1/n}## could approach anything from 1 up (although it does approach 1 ;)).

    To illustrate this: ##(1+{1 \over n})^n## approaches the mathematical constant e.
     
  8. Jul 4, 2012 #7
    I am sorry "I Like Serena", I don't understand what you mean by "n1/n could approach anything from 1 up". What ever value n has, you will be taking it to the power of 0. So it has to approach 1. I know I am wrong but I don't understand how.
     
  9. Jul 4, 2012 #8

    I like Serena

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    You might also say that whatever value (1/n) has, you will be taking infinity to the power of this number.
    So then it would have to approach infinity.
     
  10. Jul 4, 2012 #9
    That makes sense. So am I better off going with my original approach of using L'Hops rule after taking the natural log of both sides?
     
  11. Jul 4, 2012 #10

    I like Serena

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  12. Jul 4, 2012 #11
    I am not getting the right answer. (ln(3/n))/n: the derivative of the numerator is (-1/n) and the derivative of the denominator is 1. Which gives me lim = 0. Did I differentiate correctly?
     
  13. Jul 4, 2012 #12

    I like Serena

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    Actually, that looks good. :)

    Remember which limit you approached?
     
  14. Jul 4, 2012 #13
    Ah, that was the limit of natural log of the sequence. Since ln(a sub n) = 0, (a sub n) = e^0=1. Thank you so much for guiding me through this. I feel so much better.

    by the way, is there an easier way to write (a sub n)?...I'm new to this.
     
  15. Jul 4, 2012 #14

    I like Serena

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    Good! :smile:

    Try:
    Code (Text):
    ##a_n##
    That comes out as ##a_n##.

    Similarly
    Code (Text):
    $$\lim_{n \to \infty} a_n$$
    comes out as:
    $$\lim_{n \to \infty} a_n$$
     
  16. Jul 4, 2012 #15
    Got it, Thanks again
     
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