limit....
2012-10-25 7:41 am
回答 (1)
2012-10-26 8:03 am
✔ 最佳答案
(6x^3 - 13x^2 - 41x - 12) / ( x^2 - 8x + 16 )=[(x-4)(2x+3)(3x+1)] / [(x-4)^2]
=[(2x+3)(3x+1)] / (x-4)
Let y = [(2x+3)(3x+1)] / (x-4)
For x-> 4-, (x-4) -> 0-
Then y -> -inf
For x-> 4+, (x-4) -> 0+
Then y -> inf
As lim (x-> 4-) does not equal to lim (x-> 4+) so the limit (x-> 4) does not exist.
收錄日期: 2021-04-23 20:47:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121024000051KK00677