In an assignment for the course Real-Analysis I need to proof that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is not continuous in $(0, 0)$ using an Epsilon-Delta proof.
However, me and my fellow students have been trying to figure it out for some time now, but we can't...
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Hint: In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$ and
$f(x, y)=\frac{2xy}{x^2 + y^2}=\frac{2r^2\cos\theta\sin\theta}{r^2}=2\cos\theta\sin\theta$.
When $r \to 0$, the limit does not exist. Why?
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Suppose that the limit exists and it is equal to $L$.
For any $\varepsilon>0$, we can find $\delta>0$ so that $|(x,y)|<\delta$ implies that $|f(x,y)-L|<\varepsilon$.
Fix $\varepsilon>0$, but arbitrary. Find such $\delta>0$, so for $|x|<\delta$ and $y=0$ we have that $|(x,y)|<\delta$, thus $|f(x,y)-L|<\varepsilon$, but since $y=0$ we have that $f(x,y)=0$, so $|L|<\varepsilon$. Likewise, for $|x|<\sqrt{\frac{1}{2}\delta}$ we have that $|(x,x)|<\delta$, so $|f(x,x)-L|<\varepsilon$, but $f(x,x)=2x^2/(x^2+x^2)=1$, so $|1-L|<\varepsilon$.
So we have concluded that $|L|<\varepsilon$ and $|1-L|<\varepsilon$ and this is true for any $\varepsilon>0$. Is this possible for $\varepsilon=1/10$?
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Alternative approach:
Suppose that it is given that $(x,y)$ approaches $(0,0)$ along the line $y = kx$, where $k$ is a fixed positive real number. Then, you can compute the (non-zero) limit in terms of $k$. That is, the limit will vary, as $k$ varies.
Therefore, since $(x,y)$ can approach $(0,0)$ along any line $y = kx$, where (for example) $k \in \mathbb{R^+}$, and since the limit will vary as $k$ varies, it is impossible for there to be a single limit that would pertain, regardless of the value of $k$.
That is, it is impossible for there to be a single limit that would pertain, regardless of the line $(y = kx)$ along which $(x,y)$ approached $(0,0)$.
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In any neighborhood of $(0,0),$ regardless of the size of delta, there will be a point $(x,x)$ such that $f(x,x) = 1$ and there will be a point $(x,-x)$ such that $f(x,-x) = -1$
For $\epsilon = 1$ and any $\delta > 0$ and any value of $L,$ there is a point $(x,y)$ with $d((x,y),(0,0))<\delta$ and $|f(x,y) - L| > \epsilon.$
Edit: My first post only showed that $f$ cannot be continuous by declaring $f(0,0) = 0$. Now I adjusted the proof to show that $f$ is not continuous at $(0,0)$.
Let us suppose for the sake of contradiction that $f$ is continuous at $(0,0)$ and $\lim_{(x,y)\rightarrow(0,0)}f(x,y) = L$. Note that $f(x,x) = \frac{2x^2}{x^2+x^2} = 1$ and $f(2x,x) = \frac{4x^2}{4x^2+x^2} = \frac{4}{5}$. Now let $\epsilon = \frac{1}{20}$ and let $\delta > 0$ be such that $|f(x,y)-L| < \epsilon$ whenever $(x,y)$ is within $\delta$ units of $(0,0)$. Since the points $(\frac{\delta}{4},\frac{\delta}{4})$ and $(\frac{\delta}{2},\frac{\delta}{4})$ are within $\delta$ units of $(0,0)$, we see that $\frac{1}{20} > |f(\frac{\delta}{4},\frac{\delta}{4}-L| = |1-L|$ and $\frac{1}{20} > |f(\frac{\delta}{2},\frac{\delta}{4})-L| = |\frac{4}{5}-L|$. It follows that $\frac{1}{10} = \frac{1}{20}+\frac{1}{20} > |1-L|+|L-\frac{4}{5}| \ge |1-\frac{4}{5}| = \frac{1}{5}$, which yields the desired contradiction.