来自李永奎的问题
求极限(sinx)^tanx,当x趋向与π/2的时候=lim(x->Л/2)[(1/sinx)*cosx]/[-(sinx)^2-(cosx)^2/(sinx)^2]=lim(x->Л/2)1/2*sin2x=0
求极限(sinx)^tanx,当x趋向与π/2的时候
=lim(x->Л/2)[(1/sinx)*cosx]/[-(sinx)^2-(cosx)^2/(sinx)^2]
=lim(x->Л/2)1/2*sin2x
=0
1回答
2020-12-20 20:48