# To compute <munder> <mo movablelimits="true" form="prefix">lim n </munder> ( 1 +

To compute $\underset{n}{lim}\left(1+n{\right)}^{\frac{1}{\mathrm{ln}n}}$ without L'Hospital
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Aaron Everett
I always treat limits of this form by computing the limit of the logarithm, so we want to consider
$\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(1+n\right)}{\mathrm{ln}n}$
If we consider instead of the sequence the function limit
$\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(1+x\right)}{\mathrm{ln}x}$
we know that if this one exists, then it's the same of the limit of the sequence (the converse is generally not true). Now we can perform the substitution $x=1/t$ and note that
$\mathrm{ln}\left(1+1/t\right)=\mathrm{ln}\left(t+1\right)-\mathrm{ln}t$
so the limit becomes
$\underset{t\to {0}^{+}}{lim}\frac{\mathrm{ln}t-\mathrm{ln}\left(1+t\right)}{\mathrm{ln}t}=\underset{t\to {0}^{+}}{lim}\left(1-\frac{\mathrm{ln}\left(1+t\right)}{\mathrm{ln}t}\right)=1$
because the fraction is $0/\mathrm{\infty }$
Thus we have proved that
$\underset{n\to \mathrm{\infty }}{lim}\left(1+n{\right)}^{1/\mathrm{ln}n}={e}^{1}=e$