Question

limx→0 (cosx)^[1/xln(1-x)]=?

Answer 1

以下limx→0不写了

=e^{ln[(cosx)^[1/xln(1-x)]]}

其中ln[(cosx)^[1/xln(1-x)]]

=ln(cosx)/(xln(1-x)) 洛必达法则

=-tanx/[ln(1-x)-x/(1-x)] tanx与x在x->0是等价无穷小

=x/[x/(1-x)-ln(1-x)]

=x(1-x)/[x-(1-x)ln(1-x)] 洛必达法则

=(1-2x)/(2+ln(1-x))

=1/2

所以原式=e^(1/2)