# Solve a limit with the form (x^n)(e^ax) when x approaches to infinity?

1. Oct 28, 2012

### Wilfired

solve a limit with the form (x^n)(e^ax) when x approaches to infinity???

Well, my question is how to solve a limit with the form (x^n)(e^ax) when x approaches to infinity using L´Hopital rule??
I made a try, transforming the limit to (x^n)/(e^-ax), and using L´Hopital repeatedly, gives me something like this:
(nx^n-1/(-ae^ax), (n(n-1)x^n-2/(a^2e^-ax)....
So, the question, if this is correct (although i don't think so), is how i can simplify this??
If i´m wrong (which is more probably), please don't be so hard with me, hehe :)
Thanks for the answers!!

2. Oct 28, 2012

### lurflurf

Re: solve a limit with the form (x^n)(e^ax) when x approaches to infinity???

You do not need L´Hopital rule. Show (x^n)(e^ax) and (e^ax) have the same limit.

3. Oct 28, 2012

### haruspex

Re: solve a limit with the form (x^n)(e^ax) when x approaches to infinity???

Is it known that n and a are > 0?

4. Oct 28, 2012

### Wilfired

Re: solve a limit with the form (x^n)(e^ax) when x approaches to infinity???

Yes!!

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