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

  • 2
    $\begingroup$ This is so much more complicated than needed, seemingly with the goal of losing all intuition about the situation... Why not simply consider limits along the lines $x=0$ and $x=y$ and conclude right away? $\endgroup$
    – Did
    Aug 23, 2018 at 6:16
  • $\begingroup$ Because I specifically need a proof by the negation of $\epsilon-\delta$ definiton. My problem is mainly related to logic. $\endgroup$
    – user561334
    Aug 23, 2018 at 16:15

3 Answers 3


Suppose the limit is $L$.

Then there exists $\delta>0$ such that, for $0<\sqrt{x^2+y^2}<\delta$ (with $y\ne0$), $$ \left|\frac{x}{y}-L\right|<1 $$ that is, $$ L-1<\frac{x}{y}<L+1 $$ 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 $$ t=\frac{1}{3} \qquad x=\frac{1}{2}\frac{\delta}{\sqrt{1+(1/3)^2}} $$ to falsify $0<x/y<2$.

  • $\begingroup$ This is a proof by contradiction, right? I'm needing to prove directly the negation. I think I can use your answer to write it. $\endgroup$
    – user561334
    Aug 23, 2018 at 16:23
  • $\begingroup$ @dude3221 What do you mean by “direct proof” of a false statement? $\endgroup$
    – egreg
    Aug 23, 2018 at 16:51
  • $\begingroup$ Direct proof of a negation. Prove that $\exists \epsilon>0$ such that $\forall \delta > 0$ $\exists (x,y)$ such that $||(x,y)-(0,0)||<\delta$ and $|x/y - L|\geq \epsilon$. But I think is almost the same as you did by contradiciton, just need to rewrite things, right? Looks fine to me. I'll accept your answer, thanks. $\endgroup$
    – user561334
    Aug 23, 2018 at 16:58

Assume that $L\ne 1$ 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 $${\bf z}_n:=\left({1\over n},{(-1)^n\over n}\right)\qquad(n\geq1)$$ converges to $(0,0)$, but the function values $f({\bf z}_n)={\displaystyle{x_n\over y_n}=(-1)^n}$ have no limit in ${\mathbb R}\cup\{\pm\infty\}$.


You must log in to answer this question.