求极限:(x→1) lim(1-X)tg(π/2x)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上面的做法错在哪里?(x→1) lim(1-X)tg(π/2 x) =lim(1-X)*cot(π/2-π/2 x) =lim(1-X)/tan(π
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求极限:(x→1) lim(1-X)tg(π/2x)
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
上面的做法错在哪里?
(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
tanx~x 是等价无穷小 不是等价无穷大
后面的解法是对的呀...
等价无穷小用错了
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求极限:(x→1) lim(1-X)tg(π/2x)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上面的做法错在哪里?(x→1) lim(1-X)tg(π/2 x) =lim(1-X)*cot(π/2-π/2 x) =lim(1-X)/tan(π
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