lim (x^9 - y^9) / (x^5 - y^5)
x->y
係咪用得L'Hospitals' rule ? 點做?
更新1:
哦~即係唔使put 就可以用L'Hospitals' rule 啦? 如果put 左x=y+h之後 lim (h->0) [(y+h)^9-y^9)/[(y+h)^5-y^5] 仲用唔用到 L'Hospitals' rule ?
pure maths
2010-04-14 4:47 am
put x=y+h, evaluate the limit
lim (x^9 - y^9) / (x^5 - y^5)
x->y
係咪用得L'Hospitals' rule ? 點做?
lim (x^9 - y^9) / (x^5 - y^5)
x->y
係咪用得L'Hospitals' rule ? 點做?
回答 (1)
2010-04-14 5:53 am
✔ 最佳答案
Let x=y+hlim (x->y) (x^9-y^9)/(x^5-y^5)
=lim (h->0) [(y+h)^9-y^9)/[(y+h)^5-y^5]
=lim (h->0) [(y^9+9y^8h+36y^7h^2+...-y^9)/[y^5+5y^4h+10y^3h^2+...-y^5]
=lim (h->0) [(9y^8h+36y^7h^2+...)/[5y^4h+10y^3h^2+...]
=lim (h->0) [(9y^8+36y^7h+...)/[5y^4+10y^3h+...]
=(9/5)y^4
Using L'Hospitals' rule
lim (x->y) (x^9-y^9)/(x^5-y^5)
=lim (x->y) (9x^8)/(5x^4)
=(9/5)y^4
2010-04-15 14:10:25 補充：
OK
收錄日期: 2021-04-26 14:04:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100413000051KK01473