Proving partial deviatives not continous

  • Thread starter Thread starter wowowo2006
  • Start date Start date
  • Tags Tags
    Partial
wowowo2006
Messages
13
Reaction score
0

Homework Statement


f(x,y) = y^2 + (x^3)*sin(1/x) when x =/= 0
= y^2 when x = 0

i want to prove fx(x,y) is not continuous at (0,0)

Homework Equations





The Attempt at a Solution


i found when x=/=0 , fx = 3(x^2)sin(1/x) - xcos(1/x) -----eq(1)
and limit(x,y -> 0,0) eq(1) = 0 as sin and cos is bounded
and the actual fx(0,0) = limit(h->0) (f(h,0)-f(0,0))/h = lim(h->0) (h^2)*sin(1/h) = 0
it seem limfx(0,0) = fx(0,0)
so i cannot conclude that fx is not continuous at (0,0)
where did i go wrong?
 
Physics news on Phys.org
can someone offer help please?.
 
I think the partial derivative is actually continious. Graphing it shows it has no oscillation, which is a quantitative definition of continuity.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
2
Views
1K
Replies
4
Views
1K
Replies
11
Views
2K
Replies
4
Views
1K
Replies
12
Views
2K
Replies
23
Views
3K
Replies
6
Views
1K
Back
Top