# show limit of e^(1/n) / n = 0

• Feb 3rd 2010, 03:49 PM
buckeye1973
show limit of e^(1/n) / n = 0
Hi all again,

I know that this limit = 0, but I'm blanking on how to show it..

$\lim_{n\to\infty} \frac{e^{1/n}}{n} = 0$

I'm trying to manipulate it into an acceptable form for l'Hospital, but it isn't coming to me. Thanks for any help!

Brian
• Feb 3rd 2010, 03:56 PM
Calculus26
don't need L'Hopital as directly you get e^(0)/inf = 1/inf = 0
• Feb 3rd 2010, 05:19 PM
buckeye1973
Quote:

Originally Posted by Calculus26
don't need L'Hopital as directly you get e^(0)/inf = 1/inf = 0

Oh! Of course...duh..(Itwasntme) heh, thanks.
• Feb 3rd 2010, 05:31 PM
Drexel28
Quote:

Originally Posted by buckeye1973
Hi all again,

I know that this limit = 0, but I'm blanking on how to show it..

$\lim_{n\to\infty} \frac{e^{1/n}}{n} = 0$

I'm trying to manipulate it into an acceptable form for l'Hospital, but it isn't coming to me. Thanks for any help!

Brian

Quote:

Originally Posted by Calculus26
don't need L'Hopital as directly you get e^(0)/inf = 1/inf = 0

A little more rigorously note that $\left|\frac{e^{\frac{1}{n}}}{n}\right|\leqslant\fr ac{1}{n}$
• Feb 3rd 2010, 05:45 PM
TWiX
Quote:

Originally Posted by Drexel28
A little more rigorously note that $\left|\frac{e^{\frac{1}{n}}}{n}\right|\leqslant\fr ac{1}{n}$

So we will apply the sandwich theorem, right ?
• Feb 3rd 2010, 05:47 PM
Drexel28
Quote:

Originally Posted by TWiX
So we will apply the sandwich theorem, right ?

True dat.