lim1/(n^1/n)(n~无穷)=?

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lim1/(n^1/n)(n~无穷)=?
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lim1/(n^1/n)(n~无穷)=?
lim1/(n^1/n)(n~无穷)=?

lim1/(n^1/n)(n~无穷)=?
lim1/(n^1/n)(n~无穷)=1

y = 1/[n^(1/n)] = n^(-1/n)
lny = (-1/n)ln(n) ┏↓下面这两步运用洛必达法则↓┓
lim(n-->∞) lny = lim(n-->∞) -ln(n)/n = lim(n-->∞) -1/n = 0
=> lny = 0 => y = e^0 = 1
==> lim(n-->∞) 1/[n^(1/n)] = 1