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(1-x)tan(πx/2)=lim(1-x)/cot(πx/2)=lim(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
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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(1-x)tan(πx/2)=lim(1-x)/cot(πx/2)=lim(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
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(1-x)tan(πx/2)=lim(1-x)/cot(πx/2)
=lim(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
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(1-x)tan(πx/2)=lim(1-x)/cot(πx/2)=lim(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
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.