Can we check if limit exists using $\epsilon$-$\delta$ method. We can prove that limit exists using $\epsilon$-$\delta$ method. But, can we conclude a limit's existence using this method? As I saw some other questions where limit does not exist, it felt we can form a relation between $\epsilon$ & $\delta$ using inequalities. At this moment i am really confused. Some help will be appreciated.
for example I can't prove that limit does not exist for $\frac{x^2y^2}{x^4+y^4}$ using $\epsilon$-$\delta$ method.
Thanks
 A: We say that the limit of $f(x, y)$ at a point $(a,b)$ exists if and only if there is $c \in \mathbb R$ such that for all $\varepsilon > 0$ there is $\delta_\epsilon > 0$ with
$$\|(x, y) - (a, b)\|_2 \le \delta_\epsilon \Rightarrow |f(x, y) - c|\le  \varepsilon.$$
In this case we write $\lim_{(x, y) \to (a, b)} f(x, y) = c$.
Now I will show that there is no $c \in \mathbb R$ that satisfies the above formula for $(a, b) = (0, 0)$ and $f(x, y) = x^2y^2/x^4 + y^4$. We do that by contradiction by assuming that there exists $c \in \mathbb R$ with $\lim_{(x, y) \to (0, 0)} f(x, y) = c$ (we assume that $c > 0$, the other cases can be done using the same argument).
Let us fix $\varepsilon > 0$ and the associate $\delta_\epsilon > 0$. We should have that for all $(x, y)$ smaller than $\delta_\epsilon$, $|f(x, y) - c|$ is smaller than $\varepsilon$, or, in other words, $c - \varepsilon \le f(x, y) \le c + \varepsilon$. However, if you take
$$(x, y) = (\min\{\delta_\epsilon, 1/\sqrt{c + \varepsilon}\}, 0),$$
you find that $\|(x, y)\|_2 = \sqrt{x^2 + y^2} = \min\{\delta_\epsilon, 1/\sqrt{c + \varepsilon}\} \le \delta_\varepsilon$ and
$$f(x, y) = \frac{x^2y^2}{x^4 + y^4} = \frac{1}{\min\{\delta_\epsilon, 1/\sqrt{c + \varepsilon}\}^2} > c + \varepsilon$$
which is a contradition to our assumption. Therefore $f(x, y)$ diverges.
A: If $\lim_{(x,y)\to (0,0)} \frac {x^2y^2}{x^4 + y^4}$ exist there would be an $L$ so that for any $\epsilon > 0$ we could find a $\delta_\epsilon$ so that whenever $0 < d((x,y),(0,0)) < \delta$ we must have $|\frac {x^2y^2}{x^4+y^4} - L| < \epsilon$.
If that is so then for any two $(x_0,y_0),(x_1,y_1)$ where $0 < d((x_0,y_0),(0,0)) < \delta$ and $0 < d((x_1,y_1),(0,0)) < \delta$ we must have $|\frac {x_0^2y_0^2}{x_0^4+y_0^4} -\frac {x_1^2y_1^2}{x_1^4+y_1^4}|   \le d|\frac {x_0^2y_0^2}{x_0^4+y_0^4}-0| +  |\frac {x_1^2y_1^2}{x_1^4+y_1^4}-0| < 2\epsilon$.
So Contrpositively, if we can show for any $\delta > 0$ there will always be $(x_0,y_0),(x_1,y_1)$ where $|\frac {x_0^2y_0^2}{x_0^4+y_0^4} -\frac {x_1^2y_1^2}{x_1^4+y_1^4}| \ge C$ for some positive constant that will prove no limit can exist.  (Because it contradicts the paragraph above.)
So for any $\delta > 0$ we have $(0,\frac 12\delta)$ is within $\delta$ of $(0,0)$.  And $\frac {0^2\cdot (\frac 12\delta)^2}{0^4 + (\frac 12\delta)^4}=0$.
But $(\frac 12 \delta, \frac 12\delta)$ is also withing $\delta$ of $(0,0)$.  And $\frac {(\frac 12\delta)^2\cdot (\frac 12\delta)^2}{(\frac 12\delta)^4+ (\frac 12\delta)^4}= \frac {\frac 1{16}\delta^4}{\frac 18\delta^4} = \frac 12$.
So $|\frac {0^2\cdot (\frac 12\delta)^2}{0^4 + (\frac 12\delta)^4}-\frac {(\frac 12\delta)^2\cdot (\frac 12\delta)^2}{(\frac 12\delta)^4+ (\frac 12\delta)^4}|=\frac 12$ which is a positive constant.
So for any $\frac 14 > \epsilon > 0$ we can't find any $\delta$ that does what we want for any $L$.
So no limit exists.
