Proof continuity of a function in R^3

prettymidget
Messages
22
Reaction score
0

Homework Statement


Define F(x,y) = {1+x^2+y^2 when x>2^(1/2) AND y<2^(1/2)}
{1-(x^2+y^2) when x>2^(1/2) OR y>2^(1/2)}

Where in R^3 is F continuous? Prove it.

Homework Equations



definition of continuity

The Attempt at a Solution


I'm having a difficult time picturing how this function looks, but I think that F is continuous on R^3 except at (2^(1/2), 2^(1/2)).

Let e>0. To show F is continuous at (a,b) I want to show there exists a d>0 such that |(x,y)-(a,b)|<d implies |f(x,y)-f(a,b)|<e. Let d=|(a,b)-(2^(1/2), 2^(1/2))|.
 
Physics news on Phys.org
Going to epsilons and deltas is too much work. It helps to think about where the continuity condition can fail. If you're inside of one of the parts of the domain on which the function is piecewise defined, the function is obviously continuous just by looking at the definition of the function on that piece. So you're only interested in the boundary of those pieces. How can you decide (without doing too much work) whether the function is continuous on the boundary of those two parts of the domain?
 
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...
Back
Top