Limit of a sequence

  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. Tom Mattson

    Tom Mattson 5,526
    Staff Emeritus
    Science Advisor
    Gold Member

    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. LOL, thanks Tom and L'hopital

    lim ln(an+b)/n

    = lim a/(an+b)
  5. 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. HallsofIvy

    HallsofIvy 41,269
    Staff Emeritus
    Science Advisor

    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. 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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