设y＝sinx^(tanx),则lny＝tanx×ln(sinx)＝sinx/cosx×ln[1+(sinx-1)]
x→π/2时,sinx→1,ln[1+(sinx-1)]等价于sinx－1,所以
lim(x→π/2) lny
＝lim(x→π/2) tanx×ln(sinx)
＝lim(x→π/2) 1/cosx×(sinx-1) 洛必达法则
＝lim(x→π/2) cosx/(－sinx)
＝0
所以,lim(x→π/2) sinx^(tanx)＝e^(0)＝1