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

  1. Oct 28, 2012 #1
    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. jcsd
  3. Oct 28, 2012 #2

    lurflurf

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    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.
     
  4. Oct 28, 2012 #3

    haruspex

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    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?
     
  5. Oct 28, 2012 #4
    Re: solve a limit with the form (x^n)(e^ax) when x approaches to infinity???

    Yes!!
     
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