更新1:
Wan: You typed my question wrong. It is x^3y^2, not x^3+y^3
更新2:
sorry I omitted a "+". It should be x^3+y^2
Limit of a function
2013-09-18 8:26 pm
If f(x,y) = (x^3+y^2)^1/3 , how to find fx(0,0) if not using the definition of differentiation?
回答 (5)
2013-09-19 7:24 pm
✔ 最佳答案
我們用不同的方向去逼近該點時，極限都會不一樣，因此不唯一，所以不存在。圖片參考：http://imgcld.yimg.com/8/n/AA00645012/o/20130919112237.jpg
2013-09-19 11:25:55 補充：
圖片還沒出現嗎@@，好像要等一下下。
2013-09-30 5:56 am
thank you 自由自在!
2013-09-20 5:31 pm
f_x mean varying x and keeping y constant. Since y = 0, so along the x-axis
f(x,y) = (x^3 + 0)^(1/3) = x
f_x (0,0) = 1
When we consider y=mx, we are approaching (0,0) at an angle to the x-axis, i.e. both x and y are changing. This does not seem to be consistent with what f_x means.
f(x,y) = (x^3 + 0)^(1/3) = x
f_x (0,0) = 1
When we consider y=mx, we are approaching (0,0) at an angle to the x-axis, i.e. both x and y are changing. This does not seem to be consistent with what f_x means.
2013-09-18 9:40 pm
heart, 我相信 Atheist 指的 fx(0,0) 是指先take partial derivative w.r.t. x
2013-09-18 9:00 pm
這條f(x,y)應該沒有limit 吧﹗
通常一條Function 有limit 都係因為遇上0的POWER 0/或無限次方，又或者分母是0或無限，再彧者function 出現無限才會有極限出現。
所以, 這條應該沒有極限，所以
f(0,0) =(0^3+0^2)^1/3 =0
通常一條Function 有limit 都係因為遇上0的POWER 0/或無限次方，又或者分母是0或無限，再彧者function 出現無限才會有極限出現。
所以, 這條應該沒有極限，所以
f(0,0) =(0^3+0^2)^1/3 =0
收錄日期: 2021-04-23 23:26:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130918000051KK00060