求(1-x)tan(兀/2)x当x->1时极限

求(1-x)tan(兀/2)x当x->1时极限
求(1-x)tan(兀/2)x当x->1时极限

求(1-x)tan(兀/2)x当x->1时极限
解法一：原式=lim(x->1)[(1-x)sin(πx/2)/cos(πx/2)]
={lim(x->1)[sin(πx/2)]}*{lim(x->1)[(1-x)/cos(πx/2)]}
=1*{lim(x->1)[(1-x)/cos(πx/2)]}
=lim(x->1)[(1-x)/sin(π/2-πx/2)] (应用诱导公式)
=lim(x->1)[(1-x)/sin(π(1-x)/2)]
=(2/π)*lim(x->1)[(π(1-x)/2)/sin(π(1-x)/2)]
=(2/π)*1 (应用重要极限lim(z->0)(sinz/z)=1)
=2/π.
解法二：原式=lim(x->1)[(1-x)sin(πx/2)/cos(πx/2)]
={lim(x->1)[sin(πx/2)]}*{lim(x->1)[(1-x)/cos(πx/2)]}
=1*{lim(x->1)[(1-x)/cos(πx/2)]}
=lim(x->1)[(1-x)'/(cos(πx/2)'] (应用罗比达法则)
=lim(x->1)[(-1)/((-π/2)sin(πx/2))] (求导数)
=-1/((-π/2)*1)
=2/π.