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Limits Verification

  1. Jan 23, 2005 #1
    Hello all

    I encountered the following problems:

    (a) Prove that [tex] \lim_{x\rightarrow \infty}((n+1)^\frac{1}{3} - n^\frac{1}{3}) = 0 [/tex]

    Using the relation [tex] a^3 - b^3 = (a-b)(a^2 + ab + b^2) [/tex] we get [tex] \frac{((n+1)^\frac{1}{3} - n^\frac{1}{3})((n+1)^\frac{1}{3})^2 + ((n+1)^\frac{1}{3}(n^\frac{1}{3}) + (n^\frac{1}{3})^2}{(n+1)^\frac{1}{3} - n^\frac{1}{3})} [/tex]

    Hence [tex] \frac {1}{((n+1)^\frac{1}{3})^2 + ((n+1)^\frac{1}{3}(n^\frac{1}{3}) + (n^\frac{1}{3})^2} [/tex]

    and this approaches 0

    Also another problem I encountered was:

    Let [tex] a_n = \frac {10^n}{n!} [/tex]
    (a) To what limit does this converge
    (b) Is this sequence monotonic
    (c) Is it monotonic from a certain n onwards
    (d) Give an estimate of the difference between [tex] a_n [/tex] and the limit.
    (e) From what value of n onwards is this difference less than [tex] \frac {1}{100}[/tex]?

    (a)Is it valid to say that just by looking we can see that the limit converges to 0? How would you do it algebraically? In other words, how do you prove that [tex] \lim_{x\rightarrow \infty}(\frac {n!}{n^n} = 0) [/tex]
    (b&c) I think this sequence is monotonic from a certain n onwards
    (d) The estimate in the difference of the limit could be any value [tex] \epsilon [/tex]. So [tex] a_n = \frac {10^n}{n!} < \epsilon [/tex] where [tex] \epsilon = \frac {1}{100} [/tex]. IS this right? How would you actually estimate the difference?

    (e) So [tex] a_n = \frac {10^n}{n!}< \frac {1}{100} [/tex]. How would you solve
    for this?

    Thanks a lot :smile:
     
    Last edited: Jan 23, 2005
  2. jcsd
  3. Jan 24, 2005 #2
    I am not sure if I can use intuituion
     
  4. Jan 24, 2005 #3

    dextercioby

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    Science Advisor
    Homework Helper

    Use the fact that the natural logarithm and the limit commute and the Stirling's approximation

    [tex] \ln n!\sim n\ln n-n [/tex]

    for very big "n"...

    Daniel.
     
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