微積分題目

👨🏻‍🏫 學術台數學

[匿名]

2013-11-02 5:14 am

x-2
Lim ───────
x→-1 x2+4x-3

2x+4
Lim ───────
x→2 x-7

x3+3x2-10x
Lim ───────
x→2 x-2

2 - √x+4
Lim ───────
x→0 x

1 3
Lim ( ──── - ──── )
x→1 x-1 x3-1

2x+1
Lim ─────
x→∞ x2-x+1

Lim √x2-2x+10
x→∞

4-√x+15
Lim ─────
x→1 x2-1

我有打根號的都是包含整串數字喔!!例如第4題是根號裡面的x+4不是根號x加4喔!!!
還有例如第一題的X2這種就是X的平方，第三題的3X2就是3X的平方以此類推!!!

麻煩寫過程給我，越詳細越好，謝謝!!!!

更新1:

不好意思倒數第二題我打錯了應該是Lim √x2-2x+10 x→-∞ 然後我這題作法還是有點不懂可以說明一下嗎謝謝 ---------------------------------------------- 最後一題為什麼16-X-15下一步驟變成X-1不是1-X嗎?

回答 (2)

螞蟻雄兵

2013-11-02 7:38 am

✔ 最佳答案

lim(x->－1)_(x－2)/(x^2+4x－3)
=(－1－2)/(1－4－3)
=1/2

lim(x->2)_(2x+4)/(x－7)
=(4+4)/(2－7)
=－8/5

lim(x->2)_(x^3+3x^2－10x)/(x－2)
=lim(x->2)_(x^2+5)
=9

lim(x->0)_[2－√(x+4)]/x
=lim(x->0)_[2－√(x+4)]*[2+√(x+4)]/｛x[2+√(x+4)]｝
=lim(x->0)_－x/｛x[2+√(x+4)]｝
=lim(x->0)_－1/[2+√(x+4)]
=－1/(2+2)
=－1/4

lim(x->1)_[1/(x－1)－3/(x^3－1)]
=lim(x->1)_(x^2+x+1－3)/(x^3－1)
=lim(x->1)_(x^2+x－2)/(x^3－1)
=lim(x->1)_(x+2)/(x^2+x+1)
=3/4

lim(x->∞)_(2x+1)/(x^2－x+1)
=lim(x->∞)_(2x/x^2+1/x^2)/(1－x/x^2+1/x^2)
=0

lim(x->∞)_√(x^2－2x+10)
=lim(x->∞)_√(x^2－2x+1)
=lim(x->∞)_(x－1)
=∞

lim(x->1)_[4－√(x+15)]/(x^2－1)
=lim(x->1)_[4－√(x+15)]*[4+√(x+15)]/｛(x^2－1)*[4+√(x+15)]｝
=lim(x->1)_(16－x－15)/｛(x^2－1)*[4+√(x+15)]｝
=－lim(x->1)_(x－1)/｛(x^2－1)*[4+√(x+15)]｝
=－lim(x->1)_1/｛(x+1)*[4+√(x+15)]｝
=－1/[2*(4+4)]
=－1/16

2013-11-02 17:00:41 補充：
lim(x->2)_(x^3+3x^2－10x)/(x－2)
=lim(x->2)_(x^2+5x)
=14

lim(x->1)_[1/(x－1)－3/(x^3－1)]
=lim(x->1)_(x^2+x+1－3)/(x^3－1)
=lim(x->1)_(x^2+x－2)/(x^3－1)
=lim(x->1)_(x+2)/(x^2+x+1)
=3/3
=1

👍 0 👎 0

2013-11-02 11:18 am

知識長筆誤:

lim(x->2)_(x^3+3x^2－10x)/(x－2)
=lim(x->2)_(x^2+5x)
=14

lim(x->1)_[1/(x－1)－3/(x^3－1)]
=lim(x->1)_(x^2+x+1－3)/(x^3－1)
=lim(x->1)_(x^2+x－2)/(x^3－1)
=lim(x->1)_(x+2)/(x^2+x+1)
=1

👍 0 👎 0

收錄日期: 2021-04-30 18:08:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131101000016KK05330

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