问下当x趋近于π/2时,(x-π/2)tan x的极限怎么求,我大一刚学高数好迷糊那.

解法一：原式=lim(x->π/2)[(x-π/2)/cotx]
=lim(x->π/2)[(x-π/2)'/(cotx)'] (0/0型极限,应用罗比达法则)
=lim(x->π/2)[1/(-csc²x)] (求导数)
=1/(-1)
=-1；
解法二：原式=lim(y->0)[y*tan(π/2+y)] (令y=x-π/2)
=lim(y->0)(-y*coty) (应用诱导公式)
=lim(y->0)[(-cosy)*(y/siny)]
=lim(y->0)(-cosy)*lim(y->0)(y/siny)
=(-1)*1 (应用重要极限lim(z->0)(sinz/z)=1)
=-1.