limit of a 2-dimensional function I know that $\frac{2xy}{x^2+y^2}$ is not steady at $(0,0)$. But I don't understand why this criteria $\left|\frac{2xy}{x^2+y^2}-0\right|\leq|2xy|$ which converges to $0$ as $x,y\to 0$ doesn't work here, which would show the contrary. When can I use this criteria in order to show that a $2$-dimensional is steady at $(0,0)$?
 A: Converting to polar coordinates, we get $f(x,y)=\displaystyle \frac{2xy}{x^2+y^2} =$
$\displaystyle f(r,\theta)=\frac{r^2\sin(\theta)\cos(\theta)}{r^2}=\frac{1}{2}\sin(2\theta)$.
Therefore, because $\displaystyle \lim_{r \to 0} f(r,\theta)$ does not exist and depends on $\theta$, we can see that $\displaystyle \lim_{(x,y) \to (0,0)}$ does not exist either, because it approaches different values from different directions.
A: By applying the AM-GM inequality $x^2+y^2\ge 2|x||y|$, for which equality holds if and only if $x=y$, yields
$$\left|\frac{xy}{x^2+y^2}\right|\le 1$$
Since equality holds for $x=y$, we cannot claim the upper bound $2|x||y|$ as proposed in the OP.
A: 1) Consider $(x, y)$ = $(t, t)$, $t \rightarrow 0$.
2) Consider $(x, y)$ = $(t, t^2)$, $t \rightarrow 0$.
Limits : 
1) lim $(x, y) \rightarrow (0,0)$ = $1$;
2) lim $(x,y) \rightarrow (0,0)$ = $0$.
Different limits for different approaches.
Limit does not exist.
A: (1). The definition of continuity of $f(x,y)$ at $(0,0)$ is that for every $e>0$ there exists $d>0$ such that whenever $|x|<d$ and $|y|<d$ we have $|f(x,y)-f(0,0)|<e.$ Or equivalently, "whenever $|x|+|y|<d$...." or "whenever $x^2+y^2<d^2$....".
(2). No matter what value you give to $f(0,0),$ if $f(x,y)=2xy/(x^2+y^2)$ when $(x,y)\neq (0,0),$ the function $f$ will be discontinuous at $(0,0).$  For if $y=x\ne 0$ then  $f(x,y)=1.$ But if $y=2x\ne 0$ then $f(x,y)=4/5.$ So if $0\ne |x|<d/2$ then $f(x,x)=1$ and $f(x,2x)=4/5.$
(3). If $xy\ne 0$ the inequality $|\frac {2xy}{x^2+y^2}-0|\leq |2xy|$ is equivalent to $x^2+y^2\geq 1.$ To test for continuity of $f$  at $(0,0)$ you have to examine the behavior of $f(x,y)$ as $x^2+y^2$ tends to $0,$ not for $x^2+y^2\geq 1.$
(4). You may have meant to write $|\frac {2xy}{x^2+y^2}|\leq 1,$ which is true when $(x,y)\neq (0,0)$, but knowing only that a function takes values only in $[-1,1]$  tells us nothing  nothing about its continuity.
(5). However we can use this inequality to prove the continuity of (for example) $g(x,y)=\pi \sqrt {|xy^{3/2}|}\;$ at $(0,0).$ We have $g(xy)=\pi \sqrt {|xy|}\; \sqrt {|y|}.$
So $|\pi \sqrt {|xy^{3/2}|}\;\leq$ $ \pi \sqrt {\frac {x^2+y^2}{2} }\;\sqrt |y|\leq$ $ \pi \sqrt {\frac {x^2+y^2}{2}}\;\sqrt {x^2+y^2}=$ $\frac {\pi}{\sqrt 2}(x^2+y^2)^{3/2}.$
So with $e>0$ and $d= \left(\frac {e\sqrt 2}{\pi} \right)^{2/3},$ if $x^2+y^2<d^2$ then $|g(x,y)|<e.$     
