Why doesn't the limit of $(1 + x - y) / (x^2+ y^2) $ exist? How to compute the following limit:
$$\lim_{(x,y)\to (0,0)}\frac{1+x-y}{x^2+y^2}$$
?
The teacher's answers is "the limit doesn't exist". But when replace the variables, the value is 
$$\frac{1 + 0 - 0}{0^2+0^2}=\frac{1}{0}=\infty.$$
Which one is correct?
 A: The limit is $\infty$ for $(x,y)$ close to $(0,0)$ you have that $1+x-y\ge\frac12$ and so $$\frac{1+x-y}{x^2+y^2}\ge \frac{1}{2(x^2+y^2)}\ge M$$ provided $0<x^2+y^2\le \frac1{2M}$.  
A: If $|x|<1/3$ and $|y|<1/3$ then
$$\frac{1+x-y}{x^2+y^2}\geq \frac{1-|x|-|y|}{x^2+y^2}\geq \frac{1/3}{x^2+y^2}\to+\infty$$
as $(x,y)\to (0,0)$. Therefore the limit is $+\infty$.
A: plugging $$x=1/n,y=1/n$$ in the term $$\frac{1+x-y}{x^2+y^2}$$ we get $$\frac{1}{2}{n^2}$$ and this goes to infinity,therefore the Limit doesn't exist.
A: I would say that $$\frac 1 0$$ gives an $NaN$, because if we consider $$\frac x y$$ to be an operation where we count the number of times we need to subtract $y$ from $x$ to get $0$, if $y=0$ then even after an infinite number of subtractions we still have not reached $0$. Now, I would say that the limit $$\lim_{y\to0} \frac x y$$ has a result equal to $\infty$ rather than $NaN$ because we can imagine that $y$ is not quite $0$ but is infinitely close, so after an infinite number of subtractions we do eventually come to reach $0$.
Cheers!
