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Limit of a Sequence

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Rigorously show that for all x>0, the limit of {x[itex]^{1/n}[/itex]} is 1.

    2. Relevant equations



    3. The attempt at a solution

    |x[itex]^{1/n}[/itex]-1|[itex]\leq[/itex][itex]\epsilon[/itex]

    I'm not sure where to go from here...just looking for a little guidance.
     
  2. jcsd
  3. Jan 26, 2012 #2
    You might want to prove it seperately for x<1 and x>1.
    What you need to do is to find and N for every epsilon so that your statement holds for all n>N.
    You might want to get rid of the absolute signs first.
     
  4. Jan 26, 2012 #3
    Ok, for the case x>0, we have:

    x[itex]^{1/n}[/itex]-1[itex]\leq[/itex][itex]\epsilon[/itex]

    [itex]\Rightarrow[/itex]N=[itex]\frac{log(x)}{log(1+\epsilon)}[/itex]

    How does this look?
     
  5. Jan 26, 2012 #4
    You mean x>1
    Other than that, it looks fine to me. Depending of your prof (or whomever this is for), you should argue with mononicity that you're doing the right thing.
    Now only thing left is x<1
     
  6. Jan 26, 2012 #5
    Right, I meant x>1.

    For x<1, we have:

    x[itex]^{1/n}[/itex]-1[itex]\geq[/itex]-[itex]\epsilon[/itex]

    [itex]\Rightarrow[/itex]N=[itex]\frac{log(x)}{log(1-\epsilon)}[/itex]

    Is this all I need to show? Also, what did you mena that I should argue with monotonicity?

    Thanks!
     
  7. Jan 26, 2012 #6
    Well done! What I mean is that you should let the reader know what's going on and what you are doing. You need to present a reason, why the inequality holds for all n>N.
    Therefore, you should carry out your calculation step by step. Also, keep in mind that you can only apply statements to an inequality if they are monotoneous. The logarithm, for example, is such a statement, because x>y means log x > log y. Of course, multiplication with a negative number is not.
    And don't forget the x=1 case.
     
  8. Jan 26, 2012 #7
    Don't there need to be values assigned to [itex]n ?[/itex]


    Don't you need for all real numbers x > 0 and all positive integers n,

    as n --> oo, the limit of x[itex]^{1/n}[/itex] is 1?



    Or, don't you need for all positive integers n, as n --> oo,

    the limit of n[itex]^{1/n}[/itex] is 1?


    Otherwise, it appears that the fraction 1/n is just sitting
    there and not being given a value.


    ------------------------------------------------------------


    Or, have for all real numbers x, as x --> oo, that the limit of

    [itex]x^{1/x}[/itex] = 1?
     
  9. Jan 26, 2012 #8
    Here n is the variable going to infinity, while x is a positive (constant) real number.
     
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