求极限:(x→1) lim(1-X)tg(π/2x)

来源：百度知道 编辑：UC知道 时间：2024/09/22 04:10:11
求极限:(x→1) lim(1-X)tg(π/2 x)步骤要详细具体????
必须用罗必达法则(求导)

tg(π/2 x）等价于pai/2x
:(x→1) lim(1-X)tg(π/2x)
=lim(x->1)(1-x)pai/2x
=lim（x->1)pai/2x-pai/2
=pai/2-pai/2
=0

(x→1) lim(1-X)tg(π/2 x)
=lim(1-X)*cot(π/2-π/2 x)
=lim(1-X)/tan(π/2-π/2 x)
=lim(1-X)/[π/2(1-X)]
=2/π
罗必达法则
(x→1) lim(1-X)tg(π/2 x)
=(x→1) lim(1-X)/cot(π/2 x)
=(x→1) lim(-1)/[π/2(-csc^2(π/2 x))]=2/π

lim(1-X)tg(π/2 x)=lim(1-x)/ctg(π/2 x)=lim(1-x)/tgπ/2(1-x)=lim(1-x)/π/2(1-x)=2/π