# Check work on 2 variable function.

by tnutty
Tags: check, function, variable, work
 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 ?
 P: 394 You showed $$\lim_{(x,y)\to (0,0)} F(x,y)=0$$ . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).
P: 328
 Quote by Billy Bob You showed $$\lim_{(x,y)\to (0,0)} F(x,y)=0$$ . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).
I thought that given value defined in the piecewise, 1 in this case, was an arbitrary value.
So it did not matter, unless the limit of part A in the piecewise function did not exist?

P: 394

## 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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