Prove using epislon and delta that the limit of $f(x,y)$ as $(x,y)$ goes to $(0,0)$ does not exist

Question

I have seen this exemple:

$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^2 -y^2}{x^2 + y^2} $$

if $ (x,y) = (x,0) $ then the function tends to 1

if $ (x,y) = (0,y) $ then the function tends to -1

Therefore the limit does not exists.

So I tried to use Epsilon-Delta Definition to do the same, but I couldn't do it.

May someone help?

Answer 1

Asserting that the limit does not exist means that$$(\forall l\in\Bbb R)(\exists\varepsilon>0)(\forall\delta>0)\left(\exists(x,y)\in\Bbb R^2\setminus\bigl\{(0,0)\bigr\}\right):\bigl\|(x,y)\bigr\|<\delta\wedge\bigl|f(x,y)-l\bigr|\geqslant\varepsilon.$$So, let $l\in\Bbb R$. Take $\varepsilon=1$. Now, let $\delta>0$. Then:

• if $l\geqslant 0$, take $(x,y)=\left(0,\frac\delta2\right)$. Then $\bigl\|(x,y)\bigr\|<\delta$ and$$\bigl|f(x,y)-l\bigr|=|-1-l|\geqslant1=\varepsilon;$$
• if $l<0$, take $(x,y)=\left(\frac\delta2,0\right)$. Then $\bigl\|(x,y)\bigr\|<\delta$ and$$\bigl|f(x,y)-l\bigr|=|1-l|\geqslant1=\varepsilon.$$

Answer 2

I think it is easier to see in polar coordinates: $f(\delta\cos \theta,\delta \sin\theta)=\cos 2\theta.$ Choose $\epsilon=1/4.$Then, given $\textit{any}\ \delta >0\ $ and $\textit{any}\ l\in \mathbb R,$ we can always choose $\theta$ so that $|\cos 2\theta-l|>1/4.$