分类：教育/科学-理工学科-数学
问题：数学极限问题求lim&lt;x→1&gt;(2
求lim<x→1>(2-x)^[tan(π/2)*x]
回答：求lim<x→1>(2-x)^[tan(π/2)*x]=lim<x→1>[1+(1-x)]^[tan(πx/2)]=lim<x→1>{[1+(1-x)]^(1/(1-x))}^[(1-x)*tan(πx/2)]=lim<x→1>e^[(1-x)*tan(πx/2)]=e^lim<x→1>tan(πx/2)/[1/(1-x)]=e^lim<x→1>[sec^2 (πx/2)*(π/2)]/[(-1)*1/(1-x)^2*(-1)]=e^lim<x→1>[sec^2 (πx/2)*(π/2)]/[1/(1-x)^2]=e^lim<x→1>(π/2)*[(1-x)^2/c^2 (πx/2)]=e^lim<x→1>(π/2)*[-2(1-x)/-2cos(πx/2)*sin(πx/2)*(π/2)]=e^lim<x→1>2(1-x)/sin(πx)=e^lim<x→1>(-2)/[cos(πx)*π]=e^[(-2)/(-π)]=e^(2/π)