Mark M. answered 01/14/18
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Retired math prof. Very extensive Precalculus tutoring experience.
limn→∞(n1/n) has the form ∞0. This is an indeterminate form.
Let y = n1/n
ln(y) = ln(n1/n) = (1/n)ln(n) = (lnn)/n
limx→∞(lnx/x) has the form ∞/∞
By L'Hopital's Rule, limx→∞(lnx/x) = limx→∞[(1/x)/1) = 0
So, limn→∞(lny) = limn→∞(ln(n)/n)) = 0
Therefore, limn→∞(n1/n) = limn→∞(y)
= limn→∞(elny) = e0 = 1