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

  • Thread starter burhan1
  • Start date
  • #1
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Homework Statement


x^n / n! -> 0 for any value of x and n -> 0


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
118
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What have you tried to do so far?
 
  • #3
7
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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...
 
  • #4
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Homework Statement


x^n / n! -> 0 for any value of x and n -> 0
I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?
 
  • #5
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I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?
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?
 
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  • #6
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Oops yeah n -> infinity. sorry about that =/
 
  • #7
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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
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,808
933
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.
 
  • #9
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im a first year undergraduate i have nothing on me panicking like crazy xD

and that didnt help TT_TT
 
  • #10
lurflurf
Homework Helper
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126
Prove that for n large
(2/n)n<1/n!<(3/n)n
 
  • #11
209
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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→∞.
 
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  • #12
7
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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
 
  • #13
lurflurf
Homework Helper
2,426
126
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
 
  • #14
7
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no that is the next question lurf :P

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

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