变量替换t=π/2-x,原极限等价于sin(t)^t,t趋向0的极限.对该式取对数得t*ln(sin(t)),变形得ln(sint)/(1/t),根据罗比达法则t趋向0时,ln(sint)/(1/t)=-t^2/sint=-2t/cost=0,所以sin(t)^t极限为1