Advanced Multivariable Calculus / Continuity / Type-o?

  • Thread starter Jamin2112
  • Start date
  • #1
986
9

Homework Statement



I don't need to state the whole problem; it's the definitions at the beginning that are giving me trouble.

Homework Equations



So it says,

Definition: A function f(x,y) is continuous at a point (x0,y0) if f(x,y) is defined at (x0,y0), and if lim(x,y)-->(x0,y0) f(x,y)=f(x0,y0).

Definition: A function f(x,y) is discontinuous at a point (x0,y0) if it is defined at (x0,y0), and if either f(x,y) had no limiting value at (x0,y0), or if lim (x,y)-->(x0,y0) f(x,y) has no value.

The problem then gives me a function f(x,y)=(xy2-y3)/(x2+y2) and asks whether lim (x,y)-->(0,0) f(x,y) has a value.

The Attempt at a Solution



Something seems wrong about the definitions. Both of them say that f is defined at (x0,y0). But what if f isn't defined there? In the function that I'm given, plugging in 0 for x and 0 for y means diving by zero. Type-o?
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
If f(x0,y0) is undefined, then f(x,y) is discontinuous at (x0,y0). Your book may have a problem with it's phrasing of the definition of 'discontinuous'. But that doesn't mean the limit doesn't exist.
 
  • #3
986
9
But that doesn't mean the limit doesn't exist.
I know. And if I change to polar coordinates it's easy to come up with a limiting value of 0.
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
I know. And if I change to polar coordinates it's easy to come up with a limiting value of 0.
Great. That's completely correct. The limit is zero. But the function is discontinuous unless they choose to define f(0,0)=0. So we agree, right?
 

Related Threads on Advanced Multivariable Calculus / Continuity / Type-o?

Replies
2
Views
2K
  • Last Post
Replies
3
Views
526
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
2K
Replies
3
Views
2K
Replies
2
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
3K
Top