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Check work on 2 variable function. 
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#1
Sep2609, 09:04 PM

P: 328

1. The problem statement, all variables and given/known data
Determine the set of points at which the function is continuous. F(x,y) = R. where R is a piecewise function of : { x^2*y^3 / (2x^2 + y^2) ; if(x,y) != (0,0) 1 ; if(x,y) = (0,0) } Obviously, the first function is not defined at point (0,0), but to find the domain of the piecewise function, I first need to see if the first function is at least continuous. So here is my attempt at that : Let A = x^2 * y^3 / (2x^2 + y^2 ) the A is = x^2 * y^3  2x^2 + y^2 well x^2 <= 2x^2 + y^2, lets call that J so A < J * y^3 / (J) = y^3 = sqrt(y^6), and we see that this function is defined at point 0 , thus lim of A as (x,y) >(0,0) = 0. ??? So if the above is true then the peicewise function should be defined in region R^2??? I am not sure if this is correct. The book says that the answer is : { (x,y)  (x,y) != (0,0) }. I think that means the function A is not defined at 0 thus the peicewise function is not defined at point (0,0). What did I do wrong ? 


#2
Sep2709, 09:02 AM

P: 392

You showed [tex]\lim_{(x,y)\to (0,0)} F(x,y)=0[/tex] . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).



#3
Sep2709, 01:24 PM

P: 328

So it did not matter, unless the limit of part A in the piecewise function did not exist? 


#4
Sep2709, 03:02 PM

P: 392

Check work on 2 variable function.
If F(0,0) had been defined to be 0 instead of 1, then F would have been continuous everywhere.
If F(0,0) had been defined to be c, with c nonzero (c=1 is a special case), then as in your problem, F would not have been continuous at (0,0) but would be continuous everywhere else. 


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