求当x趋向二分之π,(sinx-1)tanx的极限
问题描述:
求当x趋向二分之π,(sinx-1)tanx的极限
答
属于0*∞型,变形后用罗比塔法则:
lim(x-->π/2)(sinx-1)tanx
=lim(x-->π/2)(sinx-1)/cotx
=lim(x-->π/2)cosx/[-csc^2(x)]
=lim(x-->π/2)[-sin^(x)cosx]
=0