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Homework Help: Prove the limit help

  1. Oct 31, 2004 #1
    Hi...
    I have to prove that the limit as n->oo of n^(1/n) is 1
    as n->oo (1/n)-> 0, which means that n^(1/n) ->1
    i was wondering though if you could prove this differently using the definition of the limit ( with epsilon ) or maybe take derivatives..?

    thanks
     
  2. jcsd
  3. Oct 31, 2004 #2

    Hurkyl

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    Actually, your proof is wrong. Notice that your approach would also "prove" that

    [tex]
    \lim_{n \rightarrow \infty} n
    = \lim_{n \rightarrow \infty} (n^n)^{(1/n)}
    [/tex]

    would also be 1.



    Exponents are messy things -- try applying a logarithm to convert the problem into an easier one.
     
  4. Nov 1, 2004 #3

    actualy i think his proof is correct, but your limit is incorrect let me explain:
    if you "go" with the limit into the exponent of N (you can do that because N is defined on whole R) the you get->
    [tex]
    \lim_{n \rightarrow \infty} n(1/n)
    [/tex]

    n goes to infinity and 1/n goes to 0. But you always have to remember that oo is not a number its just a symbol so you cannot say that oo*0 is 0. 0*oo can be any R number even 1. Try to think about it, its a bit difficult but in the in the end its quite understandable
     
  5. Nov 1, 2004 #4

    Hurkyl

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    You might want to go rechcek the statement of the relevant limit theorems.
     
  6. Nov 1, 2004 #5

    shmoe

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    You've missed the point of Hurkyl's example and are missing your own advice on treating indeterminate limits with care. b0mb0nika had a limit of the indeterminate form [tex]\infty^{0}[/tex]. Because it was of this form he (wrongly) concluded that the limit would be 1. Hurkyl gave a limit of the form [tex]\infty^{0}[/tex] that is easily manipulated into a limit that is clearly not 1, showing the error in b0mb0nika's approach to limits of this type.

    For the original limit try using log's as already suggested, [tex]\lim_{n\rightarrow \infty}n^{1/n}=\lim_{n\rightarrow\infty}e^{\frac{1}{n}\log{n}}[/tex]

    You should be able to find [tex]\lim_{n\rightarrow\infty}\frac{1}{n}\log{n}[/tex] without too much trouble.
     
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