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

1. Oct 29, 2011

### burhan1

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. Oct 29, 2011

### murmillo

What have you tried to do so far?

3. Oct 29, 2011

### burhan1

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. Oct 29, 2011

### Staff: Mentor

I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?

5. Oct 29, 2011

### Guidenable

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
6. Oct 30, 2011

### burhan1

Oops yeah n -> infinity. sorry about that =/

7. Oct 30, 2011

### burhan1

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. Oct 30, 2011

### HallsofIvy

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, $\sum x^n/n!$ converges to $e^x$. Since it converges, the sequence of terms, $x^n/n!$ must converge to 0. But you may not be able to use that.

9. Oct 30, 2011

### burhan1

im a first year undergraduate i have nothing on me panicking like crazy xD

and that didnt help TT_TT

10. Oct 30, 2011

### lurflurf

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

11. Oct 30, 2011

### Harrisonized

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
12. Oct 30, 2011

### burhan1

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. Oct 30, 2011

### lurflurf

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. Oct 30, 2011

### burhan1

no that is the next question lurf :P

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