function continuity (function of severable variables)

• Feb 17th 2011, 11:28 PM
Taurus3
function continuity (function of severable variables)
let f(a,b)={a^4+10a^2b^2+21b^4 if (x,y) doesn't equal (0,0)
t if (x,y) equals (0,0).

Can we find a function for t that would make this function continuous or no?
• Feb 17th 2011, 11:37 PM
Prove It
Your function is not a function of $\displaystyle x$ and $\displaystyle y$...
• Feb 18th 2011, 12:12 AM
Taurus3
I don't know how to align it. But "t if (x,y) equals (0,0)" comes under {a^4+10a^2b^2+21b^4 if (x,y) doesn't equal (0,0) within the same curly bracket.
• Feb 18th 2011, 03:27 AM
HallsofIvy
"Alignment"is not the problem. In both of your posts, you have said "if (x, y) is not 0" but have written a and b in the formula. Do you mean
"[tex]f(x, y)= x^4+10x^2y^2+21y^4[tex] if (x, y) is not (0, 0), f(0, 0)= t"?

This function is clearly continuous everywhere except possibly at (0, 0). A function is continuous at a given point if and only if the value of the function there is equal to the limit of the function as (x, y) goes to that point. This function will be continuous at (0, 0) if and only if
$t= \lim_{(x,y)\to (0, 0)} x^4+ 10x^2y^2+ 21y^4$.

What is that limit?