Vtyjkyuk

I want to show, by $epsilon-delta$ definition that $$lim_(x,y) to (0,0) fracxyneq L, forall L in mathbbR$$

(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 $|fracxy-L|geq epsilon$

Let $L in mathbbR$

Consider $epsilon = 1>0$

Let $delta>0$.
First consider $(x,y)=(0,fracdelta2)$

$||(0,fracdelta2)-(0,0)|| = ||(0,fracdelta2)|| = sqrtfracdelta2^2=|fracdelta2|=fracdelta2<delta$

$|fracxy-L| = |0-L| = |L|$

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 $|L|<1$.

$(x,y) = bigg(L,dfracLL+1bigg)$ is such that

$|fracxy-L|=epsilon$, 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

Suppose the limit is $L$.

Then there exists $delta>0$ such that, for $0<sqrtx^2+y^2<delta$ (with $yne0$),
$$
left|fracxy-Lright|<1
$$
that is,
$$
L-1<fracxy<L+1
$$
Note that $x/y$ takes on both positive and negative values in the specified range, so it's necessarily $L-1<0$ and $L+1>0$, hence $-1<L<1$. In particular, $x/y<2$.

Now it's just a matter of finding $x$ and $y$ so that $0<x^2+y^2<delta^2$ and $x/y>2$.

Choose $y=tx$, with $x>0$: we need $(1+t^2)x^2<delta^2$ and $2t<1$. So we can use
$$
t=frac13
qquad
x=frac12fracdeltasqrt1+(1/3)^2
$$
to falsify $0<x/y<2$.

Assume that $Lne 1$ and take the line $(x,x)$ to get contradiction. Next assume $L=1$ and take the line $(0,y)$ to get another contradiction.

(Concerning your attempt Did's comment says it all.

The sequence $$bf z_n:=left(1over n,(-1)^nover nright)qquad(ngeq1)$$ converges to $(0,0)$, but the function values
$f(bf z_n)=displaystylex_nover y_n=(-1)^n$
have no limit in $mathbb Rcuppminfty$.