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Define the function $f(x,y)= {{x^2 + y (x^2 + y)} \over {x^2 + y^2}}$ at $[0,0]$ so that the function would be continuous.

I need help with this calculus problem. I mean, I guess it involves some limit-calculations but I don't know how to proceed.

I tried to rewrite the function as:

$f(x,y) -1 =$ $y$$x^2 \over x^2 + y^2$

I don't see the step after this.

Help please, any hint is appreciated.

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up vote 3 down vote accepted

You have $f(x,y) = 1 - y {x^2 \over x^2+y^2}$. Note that for $(x,y) \neq (0,0)$, you have $|f(x,y)-1| \le |y|$.

This both gives you the required value and a proof of continuity.

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So, at $[0,0]$ my function value is $f(x,y)=1$. And that's it? I mean, I don't quite get that how's this a proof. Is that enough for a proof, that the expression $|f(x,y)-1| \leq |y|$ holds true for [0,0] as well? – Wanderer Jun 2 '14 at 18:10
Well, the function is continuous on $\{(0,0)\}^c$. If you define $f$ to have the value $1$ at $(0,0)$, then the above shows that it is continuous at $(0,0)$. If you define $f(0,0) = 1$, then you have $|f(0,0)-1| = 0 = |0|$, and for $(x,y) \neq (0,0)$, the above formula shows that $f$ is continuous at $(0,0)$. – copper.hat Jun 2 '14 at 18:12

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