数学9求极限
求极限
解:令t=1/x,则 原式=lim(t→0+)(π/2-arctan(1/t))/t =lim(t→0+)((1/t^2)/(1+1/t^2)) =lim(t→0+)(1/(t^2+1)) =1/(0^2+1) =1.