求(1-x)tan(兀/2)x当x->1时极限
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 01:03:14
![求(1-x)tan(兀/2)x当x->1时极限](/uploads/image/z/3710630-38-0.jpg?t=%E6%B1%82%281-x%29tan%28%E5%85%80%2F2%29x%E5%BD%93x-%3E1%E6%97%B6%E6%9E%81%E9%99%90)
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.