Studying uniform continuity on a multivariable function. Prove that the function $f(x,y)=\frac{1}{x^2+y^2}$ is not uniformly continuous over the domain:
$D$={$(x,y):x^2+(y-2)^2 <2^2$}
Using the definition:
$\forall \delta>0 , \exists \epsilon(\delta) > 0 / \forall (x_1,y_1) , (x_2,y_2) 
\in D  : 0< ||(x_1,y_1),(x_2,y_2)||<\delta \implies |f(x_1,y_1)-f(x_2,y_2)|>\epsilon$
Let :
$(x_1,y_1)=(0,\frac{\delta}{2})$
$(x_2,y_2)=(\frac{\delta}{4},0)$
Conditions on $\delta$:
$x_1^2+(y_1-2)^2<2^2 \implies 0<\delta<8$
$x_2^2+(y_2-2)^2<2^2 \implies 0<\delta<8$
$|f(x_1,y_1)-f(x_2,y_2)|=\frac{12}{\delta^2}>\epsilon$
$\forall \delta>0 , \exists \epsilon(\delta) \in ]0,\frac{12}{\delta^2}[  / \forall (x_1,y_1) , (x_2,y_2) 
\in D  : 0< ||(x_1,y_1),(x_2,y_2)||<\delta \implies |f(x_1,y_1)-f(x_2,y_2)|>\epsilon$
However I'm not sure if this is correct nor I am sure if It's possible to put conditions on $\delta$ while proving uniform continuity.
I would be grateful to whoever can point out my mistakes or whoever has a much cleaner solution to this question.
Thanks in advance.
 A: Your mistakes


*

*The order of $\epsilon$ and $\delta$ is inverted in your proof.

*You've made a typo in the coordinates of $(x_2,y_2)$.  It should be $(0,\dfrac{\delta}{4})$.


Your proof continued
The statement to be proved is
$$\exists\, \epsilon_0 > 0, \forall\, \delta > 0, \exists\, (x_1,y_1), (x_2,y_2) 
\in D  : 0< ||(x_1,y_1)- (x_2,y_2)||<\delta \implies |f(x_1,y_1)-f(x_2,y_2)|>\epsilon_0.$$
Fix $\epsilon_0 > 0$ at the beginning.
From your choice of $(x_1,y_1), (x_2,y_2)$, you have $|f(x_1,y_1)-f(x_2,y_2)|=\dfrac{12}{\delta^2}$.
$$\frac{12}{\delta^2} > \epsilon_0 \iff 0 < |\delta| < \sqrt{\frac{12}{\epsilon_0}} \tag1 \label1$$
When $\delta$ satisfies \eqref{1}, we're done.  Otherwise, for "large values" of $\delta$, since the quantifier of $(x_1,y_1), (x_2,y_2)$ is "there exists", you can choose points $(x_1,y_1), (x_2,y_2)$ closer to the origin with a smaller $\delta' < \delta$ so that $$0<||(x_1,y_1)- (x_2,y_2)||<\delta'<\delta.$$  This proves the statement for all $\delta > 0$.
A shorter answer
Take a sequence in $D$: $(x_n, y_n) = (0, \dfrac1n), n \in \Bbb{N}$.  Then
$$\lim_{n\to\infty} ||(x_n,y_n) - (x_{2n},y_{2n})|| = \lim_{n\to\infty} \frac{1}{2n} = 0,$$
but
$$\lim_{n\to\infty} ||f(x_n,y_n) - f(x_{2n},y_{2n})|| = \lim_{n\to\infty} |n^2 - (2n)^2| = \lim_{n\to\infty} 3n^2 = \infty,$$
so $f$ is not uniformly continuous on $D$.
