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

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Discussion Overview

The discussion revolves around solving a limit of the form (x^n)(e^ax) as x approaches infinity, specifically exploring the application of L'Hôpital's rule and alternative approaches to evaluate the limit.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant attempts to apply L'Hôpital's rule to the limit, transforming it to (x^n)/(e^-ax) and repeatedly differentiating, but expresses uncertainty about the correctness of their approach.
  • Another participant suggests that L'Hôpital's rule is unnecessary and proposes that (x^n)(e^ax) and (e^ax) share the same limit, implying a different method may be more suitable.
  • There is a question raised about the conditions on n and a, specifically whether they are both greater than zero, which is confirmed by another participant.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of L'Hôpital's rule, indicating a lack of consensus on the best approach to solve the limit.

Contextual Notes

The discussion assumes that n and a are greater than zero, which may influence the behavior of the limit as x approaches infinity. However, the implications of this assumption are not fully explored.

Who May Find This Useful

This discussion may be useful for students or individuals interested in limit evaluation techniques, particularly in the context of exponential and polynomial functions.

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


Is it known that n and a are > 0?
 


haruspex said:
Is it known that n and a are > 0?

Yes!
 

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