I want to show, by $\epsilon-\delta$ definition that $$\lim_{(x,y) \to (0,0)} \frac{x}{y}\neq L, \forall L \in \mathbb{R}$$
(Here I am disconsidering infinite limits)
My attempt:
We must show that for every $L$ real exists $\epsilon>0$ such that for all $\delta>0$, there is $(x,y)$ such that $||(x,y)-(0,0)||<\delta$ and $|\frac{x}{y}-L|\geq \epsilon$
Let $L \in \mathbb{R}$
Consider $\epsilon = 1>0$
Let $\delta>0$. First consider $(x,y)=(0,\frac{\delta}{2})$
$||(0,\frac{\delta}{2})-(0,0)|| = ||(0,\frac{\delta}{2})|| = \sqrt{\frac{\delta}{2}^2}=|\frac{\delta}{2}|=\frac{\delta}{2}<\delta$
$|\frac{x}{y}-L| = |0-L| = |L|$
Then if $|L|\geq1=\epsilon$, it is done.
So in this case we already found $(x,y)$ such that the limit is not a real number.
Suppose $|L|<1$.
$(x,y) = \bigg(L,\dfrac{L}{L+1}\bigg)$ is such that
$|\frac{x}{y}-L|=\epsilon$, but I am failing to show that $||(x,y)||<\delta$ in this case.
Is this making sense until here? How could I finish this proof? Thanks. Thanks