分类：教育/科学-升学入学-考研
问题：计算lim(1/x2
大家来答
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回答：lim(x->1)(1-x)tan(πx/2)
=lim(y->0)[y*tan(π/2-πy/2)]  (用y=1-x代换)
=lim(y->0)[y*ctan(πy/2)]
=lim(y->0)[y*cos(πy/2)/sin(πy/2)]
=lim(y->0){[(πy/2)/sin(πy/2)]*[(2/π)*cos(πy/2)]}
={lim(y->0)[(πy/2)/sin(πy/2)]}*{lim(y->0)[(2/π)*cos(πy/2)]}
=1*(2/π)  (应用重要极限lim(x->0(sinx/x)=1)
=2/π。