-(limx→∞(sin1/x+cos1/x)^x)^iπ的答案

lim【x→∞】[sin(1/x)+cos(1/x)]^x【y=1/x】=lim【y→0】(siny+cosy)^(1/x)=lim【y→0】e^[(1/y)ln(siny+cosy)]=e^lim【y→0】[(1/y)ln(siny+cosy)]【0/0型】=e^lim【y→0】[(cosy-siny)/(siny+cosy)]=e^[(1-0)/(0+1)]=e原式=-e^(iπ)=-(cosπ+isinπ)=1