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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

    Mark44

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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

    lurflurf

    User Avatar
    Homework Helper

    Prove that for n large
    (2/n)n<1/n!<(3/n)n
     
  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

    lurflurf

    User Avatar
    Homework Helper

    Can you prove that
    lim (a x/n)n=0?
    because
    (2/n)n<1/n!<(3/n)n
    implies
    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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