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How do i prove that the limit of x^n / n! is 0?

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    x^n / n! -> 0 for any value of x and n -> 0

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 29, 2011 #2
    What have you tried to do so far?
  4. Oct 29, 2011 #3
    I know generally how to do it i just dont get how to write it in a formal proof. like x^n is always going to be approaching infinity slower than (n!) for x > 1 and for x = 1 its just 1/n! but what about x =< -1? because this doesnt converge to any number. however (n!) will always be greater than |x| soo...

    hey wait maybe i could prove that |x^n|/(n!) -> 0 and use the absolute value rule...
  5. Oct 29, 2011 #4


    Staff: Mentor

    I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?
  6. Oct 29, 2011 #5
    I think so too. The limit of xn/n! as n--->0 is 1, not 0 as x0=1 and 0!=1.

    As for xn/n! as n--->∞, try expanding the numerator and denominator of the limit. What does xn look like? What does n! look like?
    Last edited: Oct 29, 2011
  7. Oct 30, 2011 #6
    Oops yeah n -> infinity. sorry about that =/
  8. Oct 30, 2011 #7
    well x^n looks like a hell lotta x's multiplied together and n! looks like n(n-1)(n-2)...! and i realise that for any value of x, there exists n s.t n is wayyyyyy greater than x so n! will be wayyy greater than x^n but how do i prove itt... :S
  9. Oct 30, 2011 #8


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    Science Advisor

    It's impossible to tell you how to prove this without knowing what you have available! In my oppinion, the simplest way is to note that the series, [itex]\sum x^n/n![/itex] converges to [itex]e^x[/itex]. Since it converges, the sequence of terms, [itex]x^n/n![/itex] must converge to 0. But you may not be able to use that.
  10. Oct 30, 2011 #9
    im a first year undergraduate i have nothing on me panicking like crazy xD

    and that didnt help TT_TT
  11. Oct 30, 2011 #10


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    Homework Helper

    Prove that for n large
  12. Oct 30, 2011 #11
    Show that [x^(n+1)/(n+1)!] < [x^n/n!] when n>>x.
    Then [x^(n+1)/(n+1)!] / [x^n/n!] = x/(n+1) = 0 when n→∞.
    Last edited: Oct 30, 2011
  13. Oct 30, 2011 #12
    ive already tried those but we dont have sufficient knowledge to use that stuff yet.

    heres what im trying:
    Prove that a_n = (x^n/n!) -> 0 for all x

    Choose ε > 0. Then there exists for all N ,which belongs to natural numbers, n>N and |a_n| < ε
    so, |(x^n/n!)| < ε => |x^n| < n!*ε (since n! is positive the absolute sign is not significant?)
    so |x|/[ε^(1/n)] < [n!^(1/n)].

    so choose any natural number N s.t. N > |x|/[ε^(1/n)]

    If n>N then x^n/n! < x^N/N! < ε

    which means a_n -> 0.

    is this correct in any sense? =S
  14. Oct 30, 2011 #13


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    Homework Helper

    Can you prove that
    lim (a x/n)n=0?
    xn/n!~(a x/n)n
  15. Oct 30, 2011 #14
    no that is the next question lurf :P

    i've done it guys. thanks anyway for the help!
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