I have the following function:
$$f:\mathbb{R^{2}}\to \mathbb{R},\;\;f(x,y):=\frac{x-y}{x^{2}+y^{2}},\;\;(x,y)\neq (0,0).$$
I would like to test whether its limit exists at $(0,0)$.
$$\lim_{x\to 0}\left(\lim_{y\to 0}\frac{x-y}{x^{2}+y^{2}}\right) = \lim_{x\to 0}\left(\frac{x}{x^{2}}\right) = \lim_{x\to 0}\frac{1}{x} = +\infty$$
$$\lim_{y
\to 0}\left(\lim_{x\to 0}\frac{x-y}{x^{2}+y^{2}}\right) = -\lim_{y\to 0}\left(\frac{y}{y^{2}}\right) = -\lim_{y\to 0}\frac{1}{y} = -\infty$$
The divergent nature of the two limits indicates that the limit
$$\lim_{(x,y)\to (0,0)}f(x,y)$$
does not exist.
It would have been sufficient to show that at least one of the first two limits diverge.
Best Answer
First of all limit not existing finitely and limit not existing are two separate things.
Double limit or usual Limit of a multivariale function is $\lim_{(x,y)\to (0,0)}$ and it is not the same as iterated limit $\lim_{x\to 0}(\lim_{y\to 0})$ . Please read this wiki
In this case the limit itself does not exist. The way you have shown is not complete as you have shown that the ITERATED limits do not exist. However the existence of iterated limits is neither a necessary nor a sufficient condition for the double limit or simply the limit of a multivariable function to exist.
example:- $x\sin(\frac{1}{y})$ . Here the double limit exists but one of the iterated limit does not.
Following the wiki : - $\frac{xy}{x^{2}+y^{2}}$ has it's iterated limits equal to $0$ but the double limit does not exist. It is because if we take approach through the path $y=mx$ we get $\lim_{x\to 0}\frac{mx^{2}}{(1+m^{2})x^{2}}=\frac{m}{1+m^{2}}$ which is different for different values of $m$. But limit if it exists MUST be unique. Hence it does not exist.
So what you need to do is the following:-
You can prove that if the double limit of a multivariable function $F(x,y)$ exists then for any continuous functions $f(x)$ and $g(y)$ such that $\lim_{x\to 0}f(x)=0$ and $\lim_{y\to 0}g(y)=0$ , you will have
$$\lim_{(x,y)\to (0,0)} F(f(x),g(y))$$ must exist.
So here you take $f(x)=0$ and $g(y)=y$ to show that you get $\lim_{x\to 0}\frac{1}{x}$ which does not exists. (It is not $\infty$ as you said in your question as the left hand limit and right hand limit tend to positive and negative infinities). And here this suffices to say that the Double limit does not exist. But you can go a step further and show for $g(y)=0$ and $f(x)=x$ too.