微積分題目~~請高手們求解答!!

2010-07-11 11:39 am
這是大學一年級的微積分題目~
由於還需要算式~
請高手們~幫幫我^^"


一、 f(x)=lxl / x
lim x->0 f(x)=?


二、 lim x->3 √x - √3 / x - 3 =?


3 2
三、 lim x->2 x - 27 / x - 2x - 3 =?



請各位高手們幫幫我~~~拜託了^^"
答案和算式都要~~拜託各位高手了...

回答 (2)

2010-07-11 1:11 pm
✔ 最佳答案

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http://i841.photobucket.com/albums/zz333/tammie_0666/2010-7-11.jpg?t=1278796158

2010-07-12 02:34:10 補充:
三.
lim(x→3) (x³ - 27)/(x² - 2x - 3)
= lim(x→3) (x - 3)(x² + 3x + 9)/(x - 3)(x + 1)
= lim(x→3) (x² + 3x + 9)/(x + 1)
= (3² + 3(3) + 9)/(3 + 1)
= 27/4 ...... (答案)
參考: andrew, andrew
2010-07-11 2:58 pm
第1題

右極限 (x→0+)lim lxl/x=x/x=1

左極限 (x→0-)lim lxl/x=-x/x=-1

右極限不等於左極限

所以(x→0)lim lxl/x的極限值不存在

第2題

(x→3)lim(√x - √3) / (x - 3 )

=(x→3)lim(√x - √3)*(√x + √3) / (x - 3 )(√x +

√3)

=(x→3)lim(x - 3) / [(x - 3 )(√x+ √3)]

=(x→3)lim 1 / (√x+ √3)

= 1 / (√3+ √3)=1/2√3=√3/6

第3題

(x→2)lim (x - 27) / (x - 2x - 3 )

=(2-27)/(2-4-3)=-25/(-5)=5







2010-07-11 23:16:34 補充:
題目應該是

limX->3 (x^3 - 27) / (x ^2- 2x -3) =?

=limX->3 (x-3)(x^2+3x+9)/(x-3)(x+1)

=limX->3 (x^2+3x+9)/(x+1)

=(9+9+9)/(3+1)=27/4

2010-07-11 23:16:55 補充:
題目應該是

limX->3 (x^3 - 27) / (x ^2- 2x -3) =?

=limX->3 (x-3)(x^2+3x+9)/(x-3)(x+1)

=limX->3 (x^2+3x+9)/(x+1)

=(9+9+9)/(3+1)=27/4


收錄日期: 2021-04-13 17:21:34
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