解法一：∵lim(x->π/2)[(sinx-1)tanx]
=lim(x->π/2){[(sinx-1)/cosx]sinx}
=lim(x->π/2)[(sinx-1)/cosx]*lim(x->π/2)(sinx)
=lim(x->π/2){[sin(x/2)-cos(x/2)]/[cos(x/2)+sin(x/2)]}*1
=0*1
=0
lim(x->π/2){(sinx)^[1/(sinx-1)]}
=lim(x->π/2){(1+sinx-1)^[1/(sinx-1)]}
=e (应用特殊极限lim(x->0)[(1+x)^(1/x)]=e)
∴原式=lim(x->π/2)[(sinx)^tanx]
=lim(x->π/2)【(sinx)^{[1/(sinx-1)]*[(sinx-1)tanx]}】
=【lim(x->π/2){(sinx)^[1/(sinx-1)]}】^{lim(x->π/2）[(sinx-1)tanx]}
=e^{lim(x->π/2)[(sinx-1)tanx]}
=e^0
=1.
解法二：原式=lim(x->π/2)[(sinx)^tanx]
=lim(x->π/2){e^[tanx*ln(sinx)]}
=e^{lim(x->π/2)[tanx*ln(sinx)]}
=e^{lim(x->π/2)[ln(sinx)/cotx]}
=e^[lim(x->π/2)（-cotx/csc²x)]
=e^[lim(x->π/2)（-sinx*cosx)]
=e^0
=1.