Proof continuity of a function in R^3

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SUMMARY

The function F(x,y) is defined piecewise in R^3, with F(x,y) = 1+x^2+y^2 for x>√2 and y<√2, and F(x,y) = 1-(x^2+y^2) for x>√2 or y>√2. The function is continuous everywhere in R^3 except at the point (√2, √2). To prove continuity at a point (a,b), one must show that for every ε>0, there exists a δ>0 such that if |(x,y)-(a,b)|<δ, then |F(x,y)-F(a,b)|<ε. The continuity condition primarily fails at the boundary between the two defined regions.

PREREQUISITES
  • Understanding of piecewise functions
  • Knowledge of the definition of continuity in multivariable calculus
  • Familiarity with ε-δ proofs
  • Basic concepts of limits in R^3
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  • Study the properties of piecewise functions in calculus
  • Learn about ε-δ definitions of continuity in multivariable contexts
  • Explore continuity and limits in R^3
  • Investigate boundary behavior of functions in piecewise domains
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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))|.
 
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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?
 

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