# Is this correct? A limit

• May 11th 2008, 04:57 AM
Mathstud28
Is this correct? A limit
I was just wondering while in the shower if the following limit

$\lim_{x\to{\infty}}\frac{e^{x}}{x!}=0$

Can be done another way...Please...Please don't put down the ways that are commonly seen...I know that

$x!$ increases much faster than $a^{x}$ thus the limit is 0...but does this method work?

For aesthetics n will denote ∞

So rewriting

$\lim_{x\to{\infty}}\frac{e^{x}}{x!}$

as

$\lim_{x\to\infty}\frac{1+x+\frac{x^2}{2!}+\text{.. .}+\frac{x^n}{n!}}{x(x-1)(x-2)\text{...}(x-(n-1))(x-n)}$

Factoring out an $x^{n}$ from top and bottom we get

$\lim_{x\to\infty}\frac{x^{n}\bigg[\frac{1}{x^{n}}+\frac{1}{2!x^{n-1}}+\frac{1}{3!x^{n-3}}+...+\frac{1}{n!}\bigg]}{x^n(1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{n}{x})}$

Cancelling we would get

$\lim_{x\to\infty}\frac{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{x^{n-2}2!}+...+\frac{1}{n!}}{(1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{n}{x})}$

Which would be equilvalent to

$\frac{0+0+0...+0}{(1-0)(1-0)...(1-0)}=\frac{0}{1}=0$

Is that mathematically sound?
• May 11th 2008, 05:25 AM
PaulRS
But even the series you are using tells you the result.

Since it converges for all x, we have that $\frac{x^n}{n!}\rightarrow{0}$ for all x

(be careful with what you write, $e^x=1+\frac{x^1}{1!}+\frac{x^2}{2!}+...$ and $n!=1\cdot{2}\cdot{...\cdot{n}}$)
• May 11th 2008, 08:16 AM
CaptainBlack
Quote:

Originally Posted by Mathstud28
I was just wondering while in the shower if the following limit

$\lim_{x\to{\infty}}\frac{e^{x}}{x!}=0$

Can be done another way...Please...Please don't put down the ways that are commonly seen...I know that

$x!$ increases much faster than $a^{x}$ thus the limit is 0...but does this method work?

For aesthetics n will denote ∞

So rewriting

$\lim_{x\to{\infty}}\frac{e^{x}}{x!}$

as

$\lim_{x\to\infty}\frac{1+x+\frac{x^2}{2!}+\text{.. .}+\frac{x^n}{n!}}{x(x-1)(x-2)\text{...}(x-(n-1))(x-n)}$

Factoring out an $x^{n}$ from top and bottom we get

$\lim_{x\to\infty}\frac{x^{n}\bigg[\frac{1}{x^{n}}+\frac{1}{2!x^{n-1}}+\frac{1}{3!x^{n-3}}+...+\frac{1}{n!}\bigg]}{x^n(1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{n}{x})}$

Cancelling we would get

$\lim_{x\to\infty}\frac{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{x^{n-2}2!}+...+\frac{1}{n!}}{(1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{n}{x})}$

Which would be equilvalent to

$\frac{0+0+0...+0}{(1-0)(1-0)...(1-0)}=\frac{0}{1}=0$

Is that mathematically sound?

You want:

$L=\lim_{x\to{\infty}}\frac{e^{x}}{x!}$

What you have is not a rewrite of the given limit but a new limit:

$L_n=\lim_{x\to\infty}\frac{1+x+\frac{x^2}{2!}+\tex t{...}+\frac{x^n}{n!}}{x(x-1)(x-2)\text{...}(x-(n-1))(x-n)}$

This is a truncating the top and bottom of expansions of the terms in the original limit after $n+1$ terms so at best you have shown that the limit of $L_n$ the ratio of the truncations is zero.

So now how do you relate $L_n$ to $L$?

Answer that question satisfactorily and you will have a valid demonstration.

(use of one of Stirlings asymtotic formula for $x!$ is nicer as it give the asymtotic form for the ratio for large $x$)

RonL
• May 11th 2008, 04:41 PM
Mathstud28
Quote:

Originally Posted by CaptainBlack
You want:

$L=\lim_{x\to{\infty}}\frac{e^{x}}{x!}$

What you have is not a rewrite of the given limit but a new limit:

$L_n=\lim_{x\to\infty}\frac{1+x+\frac{x^2}{2!}+\tex t{...}+\frac{x^n}{n!}}{x(x-1)(x-2)\text{...}(x-(n-1))(x-n)}$

This is a truncating the top and bottom of expansions of the terms in the original limit after $n+1$ terms so at best you have shown that the limit of $L_n$ the ratio of the truncations is zero.

So now how do you relate $L_n$ to $L$?

Answer that question satisfactorily and you will have a valid demonstration.

(use of one of Stirlings asymtotic formula for $x!$ is nicer as it give the asymtotic form for the ratio for large $x$)

RonL

Are you saying they are not the same because there are $n+1$ terms in the expansion? If so then the result would yield the same thing if you pulled out $x^{n+1}$ and continued.

And Would the two limits relate by $0\leq{L}\leq{L_n}$ or in other words by the fact that we showed that $L_n=0$ we have $0\leq{L}\leq{0}$ Thus by the squeeze theorem $L=0$?
• May 11th 2008, 06:59 PM
CaptainBlack
Quote:

Originally Posted by Mathstud28
Are you saying they are not the same because there are $n+1$ terms in the expansion? If so then the result would yield the same thing if you pulled out $x^{n+1}$ and continued.

And Would the two limits relate by $0\leq{L}\leq{L_n}$ or in other words by the fact that we showed that $L_n=0$ we have $0\leq{L}\leq{0}$ Thus by the squeeze theorem $L=0$?

I am not saying they are not the same, just that you have not shown that they are the same. The missing step to show that they are the same looks to be a complicated as evaluating the original limit.

For fixed x the largest n=x, so arbitary values for n are not available for a double limiting process.

RonL
• May 11th 2008, 07:00 PM
CaptainBlack
Quote:

Originally Posted by Mathstud28

And Would the two limits relate by $0\leq{L}\leq{L_n}$ or in other words by the fact that we showed that $L_n=0$ we have $0\leq{L}\leq{0}$ Thus by the squeeze theorem $L=0$?

No, at least when $n=x$; $0\leq{L_n}\le L$

RonL
• May 11th 2008, 07:00 PM
Mathstud28
Quote:

Originally Posted by CaptainBlack
I am not saying they are not the same, just that you have not shown that they are the same. The missing step to show that they are the same looks to be a complicated as evaluating the original limit.

For fixed x the largest n=x, so arbitary values for n are not available for a double limiting process.

RonL

Thanks very much Captain Black! My problem is that since I have not yet taken any formal classes on this I am not accustomed to having to prove steps I take to be assumed