洛必达求极限

Question

洛必达求极限

Answer 1

(lnx)^[1/(x-1)]=e^[(lnlnx)/(x-1)]=e^(1/xlnx)
=e^(-1/x)=1
sin²（x-1)/(x²+ax+b)=2sin(x-1)cos(x-1)/(2x+a)
=2cos2(x-1)/2=1
u=arcsinx,(arcsinx)^tanx=u^(tansinu)=e^(lnu/tansinu)
=e^(cosucos²sinu/u)=e^(-sinucos²sinu-2cosucossinu)=e^-2