Evaluating Limit of Expanded e^x: 0?

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    E^x Limit
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Homework Statement



Typeset math: Evaluate the limit as n approaches infinity of x^n over e^x. Below, applying L'Hospital's rule n times gives the limit as n approaches infinity of x^n over e^x equals the limit as n approaches infinity of n! over e^x.


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


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?
 


This question is actually a part of another one.

Handwritten integration by parts: I = integral from 0 to infinity of e^(-x) x^n dx = -(x^n e^(-x)) from 0 to infinity + n I_(n-1), followed by a note that x^n e^(-x) tends to 0 as n tends to infinity.


Got any idea now?
 


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


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