Question

\[\lim_{n \to \infty} \frac{x^n}{n!} \text{ for } x >1\]

Answer 1

a hint would be perfect :-D

Answer 2

do you know the answer?

Answer 3

0

Answer 4

n! i believe is domineering

Answer 5

log lele

Answer 6

x.x.x.x.../1.2.3.4...

Answer 7

hmm differentiation wrt n?

Answer 8

take log

Answer 9

i am :-D him

Answer 10

sorry, no l'hopital. Instead, let N = [|x|] + 1. Then for every n > N
|x/n| < 1.

Using this you can show the product must go to zero.

Answer 11

\[\frac{x^{n+1}}{(n+1)!}*\frac{n!}{x^n}\]
\[\frac{x^{n+1}}{(n+1)*n!}*\frac{n!}{x^n}\]
\[\frac{x}{(n+1)}\]

n+1 takes the lead no matter what x you can think of

Answer 12

but why'd you take N=[|x|] +1james

Answer 13

i.e., the n-th term for n > N is

\[ |a_n| = \frac{|x|^N}{N!} \frac{|x|^{n-N}}{n!/(N-n)!} < \frac{|x|^N}{N!} \frac{|x|^{n-N}}{N^{n-N}} \]

Answer 14

Now, show the the RHS --> 0, which it will because |x/N| < 1.
That is why we chose N = [|x|]+1

Answer 15

correction: that middle fraction should be
\[ \frac{|x|^{n-N}}{n!/N!} \]

Answer 16

Now, knowing that for |a| < 1, \[ \lim_{n \rightarrow \infty} a^n = 0 \]
it follows that
\[ \lim_{n \rightarrow \infty} |x/N|^{n-N} = 0 \]

Answer 17

I need to work on it, thanks btw