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

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To solve the limit of the form (x^n)(e^ax) as x approaches infinity, L'Hôpital's rule may not be necessary. Instead, it's suggested to demonstrate that both (x^n)(e^ax) and e^ax have the same limit. The discussion emphasizes that both n and a should be greater than zero for the limit to be evaluated correctly. The transformation of the limit to (x^n)/(e^-ax) is noted, but the simplification process is questioned. Ultimately, understanding the behavior of exponential functions compared to polynomial functions is crucial in this limit evaluation.
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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