当x趋于0+时,(ln1/x)^x的极限怎么求?用洛比达法则

Question

当x趋于0+时,(ln 1/x)^x的极限怎么求?用洛比达法则

Answer 1

=e^lim(x->0+) ln(ln(1/x))/(1/x)
令t=1/x
=e^lim(t->+∞) ln(lnt)/t
=e^lim 1/(tlnt)
=e^0
=1