lim(x→0)(cos2x)^(x^(1/2))
设 y=(cos2x)^(x^(1/2))
lny=x^(1/2)*ln(cos2x)
lim(x→0) lny=0
所以 lim(x→0)(cos2x)^(x^(1/2))=e^0=1

limx(x→∞)(lnx)^[1/(x-1)]
设y=(lnx)^[1/(x-1)]
lny=[1/(x-1)]*lnx
limx(x→∞)lny=limx(x→∞)[1/(x-1)]*lnx
=limx(x→∞) lnx/(x-1)=limx(x→∞) =limx(x→∞) 1/x=0
所以 limx(x→∞)(lnx)^[1/(x-1)]=e^0=1第二题不是应该是lim(lnx)^[1/(x-1)]=e^lim lnlnx*[1/(x-1)]limx(x→∞)(lnx)^[1/(x-1)]设y=(lnx)^[1/(x-1)]lny=[1/(x-1)]*lnlnxlimx(x→∞)lny=limx(x→∞)[1/(x-1)]*lnlnx=limx(x→∞) lnlnx/(x-1)=limx(x→∞) (1/x/(lnx))=limx(x→∞) 0/∞=0所以 limx(x→∞)(lnx)^[1/(x-1)]=e^0=1