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Limit of a sequence

  1. Jun 26, 2003 #1
    It isn't a homework problem but I think I better post it here instead of Mathematics forum, since it belongs to "exam help".

    Prove that for any positive real numbers a and b,
    lim [(an+b)1/n-1] = 0

    I don't need to use things like |a-b|<epsilon. A simple way will do. I know it's an easy question but I don't know where to start. Could someone please help.
  2. jcsd
  3. Jun 26, 2003 #2

    Tom Mattson

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    This one just screams "L'Hopital!"

    First, rearrange it to:


    Then take the natural log of both sides to get:

    lim ln(an+b)/n=0

    This goes to &infin;/&infin;, which is an indeterminate form and ripe for L'Hopital's rule.
  4. Jun 26, 2003 #3
    LOL, thanks Tom and L'hopital

    lim ln(an+b)/n

    = lim a/(an+b)
  5. Jun 27, 2003 #4
    Oh sorry, I forgot to mention
    is a sequence, not a function. I think L'hopital's rule applies to differentiable functions only.

    Perhaps I better rephase the question a bit.
    A sequence {an} is defined by (an+b)1/n-1
    Prove that
    lim (an+b)1/n-1 = 0
    (a and b are real numbers and n is a positive integer)
  6. Jun 27, 2003 #5


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    It is true that L'hopital's rule applies to functions rather than sequences.

    However, IF we can convert a sequence an to a function f(x) (we can't if the sequence involves things like n! or "floor" or "ceiling" that can't be written simply as a continuous function), then f(x)-> L, an-> L. The other way doesn't necessarily work- the function might not have a limit, depending on how it is defined for non-integer values.
  7. Jun 27, 2003 #6
    So we can treat a sequence as a function if it is an "elementary" one like the one I posted, and can apply L'hopital's rule, is it correct?
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