# Proving multivariable limit doesn't exist using $\epsilon$ - $\delta$ definition?

How would you use first principles to prove that a limit doesn't exist? Let $f: D \rightarrow \mathbb{R}$ be a function of two variables where $D \subseteq \mathbb{R}^2$.

We say that the limit $$\lim_{(x,y) \rightarrow (a,b)} f(x,y) = L$$exists if and only if

$$(\forall \epsilon >0)(\exists \delta(\epsilon) >0)(\forall (x,y) \in D)(|| (x,y) - (a,b)|| < \delta(\epsilon) \Rightarrow |f(x,y) - L| < \epsilon)$$

To prove that this is not that case, we can verify the negation

$$\neg (\forall \epsilon >0)(\exists \delta(\epsilon) >0)(\forall (x,y) \in D)(|| (x,y) - (a,b)|| < \delta(\epsilon) \Rightarrow |f(x,y) - L| < \epsilon)$$

$$\iff (\exists \epsilon >0) \neg (\exists \delta(\epsilon) >0)(\forall (x,y) \in D)(|| (x,y) - (a,b)|| < \delta(\epsilon) \Rightarrow |f(x,y) - L| < \epsilon)$$

$$\iff (\exists \epsilon >0) (\forall \delta(\epsilon) >0) \neg (\forall (x,y) \in D)(|| (x,y) - (a,b)|| < \delta(\epsilon) \Rightarrow |f(x,y) - L| < \epsilon)$$

$$\iff (\exists \epsilon >0) (\forall \delta(\epsilon) >0) (\exists (x,y) \in D) \neg (|| (x,y) - (a,b)|| < \delta(\epsilon) \Rightarrow |f(x,y) - L| < \epsilon)$$

$$\iff (\exists \epsilon >0) (\forall \delta(\epsilon) >0) (\exists (x,y) \in D)(|| (x,y) - (a,b)|| < \delta(\epsilon) \land |f(x,y) - L| \geq \epsilon)$$

However, I'm not exactly sure how you would apply this. The particular example I'm doing is proving that $$\lim_{(x,y) \rightarrow (0,\ n\pi)}{\frac{1}{x^2+\sin^2y}}$$ doesn't exist where $n \in \mathbb{Z}$. In this case, we need to show that $$(\exists \epsilon >0) (\forall \delta(\epsilon) >0) (\exists (x,y) \in \mathbb{R}^2 \backslash \{(0, n\pi)\}) \Big(\sqrt{x^2 + (y-n\pi)^2} < \delta(\epsilon) \land \Big|\frac{1}{x^2 + \sin^2y} - L\Big| \geq \epsilon\Big)$$

I'm not sure how to proceed from here. I know that you can take the limit along two different paths and create a contradiction, but I was wondering how you could prove the limit doesn't exist by verifying the condition above.

Note that the assertions $$({\rm a}):\quad\lim_{x\to\xi} f(x)=L,\qquad ({\rm b}): \quad \lim_{x\to\xi} f(x)\ {\rm exists}$$ are not the same. The negation of (a) is given by $$\exists \epsilon>0\quad \forall \delta>0 \quad \exists x\ :\qquad |x-\xi|<\delta\quad \wedge\quad |f(x)-L|\geq\epsilon\ ,$$ whereas the negation of (b) reads as follows: $$\forall L\in{\mathbb R}\quad \exists \epsilon>0\quad \forall \delta>0 \quad \exists x\ :\qquad |x-\xi|<\delta\quad \wedge\quad |f(x)-L|\geq\epsilon\ .$$ But having learnt various things about limits we don't have to go through four nested quantifiers in order to deal with the proposed example. In order to show that the limit does not exist it suffices to remark that $$f(x,y):={1\over x^2+\sin^2 y}$$ is unbounded in the neighborhood of any point $(0,n\pi)$, as is exemplified by looking at the values $f(x,n\pi)={1\over x^2}$ when $x\to0$.
Consider $a_m = (1/m,n\pi)$. Then $f(a_m)=m^2\rightarrow\infty$. Therefore no matter what $L$, $\epsilon$ and $\delta$ are, you can find $m$ and ${m'}$ s.t. $|f(a_m)-f(a_{m'})|$ is greater than $\epsilon$ while $a_m$ and $a_{m'}$ are within $\delta$.