Question

求lim(x->1)(1-X)tan(πx/2)的极限

因为当x→1时，cot(πx/2)=tan(π/2-πx/2)=tan[(1-x)π/2]～(1-x)π/2
所以lim<x→1>(1-x)tan(πx/2)=lim<x→1>(1-x)/cot(πx/2)
=lim<x→1>(1-x)/[(1-x)π/2]=2/π

解释下这个cot(πx/2)=tan(π/2-πx/2)=tan[(1-x)π/2]～(1-x)π/2

洛必达法则还没有学

Answer 1

解：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。

Answer 2

tan（π/2+α）=－cotα
tan（π/2－α）=cotα
令α=(πX/2)，则cot(πx/2)=tan(π/2-πx/2)

Answer 3

tanx在x趋向0时等价于x

Answer 4

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