Evaluating Limit of Expanded e^x: 0?

zorro · Dec 19, 2010

Homework Statement

$&space;}\frac{x^{n}}{e^{x}}&space;=&space;\lim_{n&space;\to&space;\infty&space;}\frac{n!}{e^{x}}.gif$

I tried expanding e^x and evaluated the limit as 1.
The answer given is 0.

Char. Limit · Dec 19, 2010

How odd. The answer actually depends on the values of x... try a couple of values of x and you'll see what I mean. Might I recommend .75 and 2?

zorro · Dec 19, 2010

This question is actually a part of another one.

$space;Hospitals&space;rule,&space;n&space;times&space;for&space;}\&space;\frac{x^{n}}{e^{x}}.....gif$

Got any idea now?

Char. Limit · Dec 19, 2010

Either there's a typo or you misread it. It should be as x\to\infty. Then it makes sense, especially since you're integrating over dx, not over dn.

zorro · Dec 19, 2010

Oh yes, that must be a typo. I got it now.
Thanks!

hunt_mat · Dec 19, 2010

To compute the limit as x\rightarrow\infty, set x=1/y amd examine the limit as y\rightarrow 0