Limit of a function with two variables: when do we stop looking for another value? So for instance we have this limit of a function $\displaystyle\lim_{(x,y)\to(0,0)}{xy\over \sqrt{2x^2+y^2}}$, and the function isn't continuous at the point $(0,0)$. Now we can try to find the limit, and there's a possibility that it doesn't exist in case we find that the function approaches different values from different paths. Is that correct? 
So for instance I tried approaching the point from y-axis ($x=0$) and x-axis ($y=0$), as well as $y=x$. The limit for all these is $0$. So it seems like the limit of the function as $(x,y)$ approach $(0,0)$ is $0$, but how do I know I haven't missed some other path, from which we might have approached different value?
 A: One can use paths, or particular sequences $(x_n,y_n)\to(0,0)$, in order to prove that such a limit does not exist. Producing  a single path or sequence where the limit does not exist, or  two different paths or sequences with different limits of the expression in question closes the case.
But in the case at hand you (correctly) conjecture that the limit does exist, and you want to prove this. Under these circumstances it is of no help considering paths: You would have to prove that the limit is the same for all paths, but there are just to many of them. In addition I'm sure you can't even imagine how complicated a path towards $(0,0)$ can look like.
Instead we have to use the actual definition of convergence in this case. This definition works with $\epsilon$, $\delta$, and neighborhoods. Fortunately you have a definite idea what the limit is, namely $0$. You therefore have to prove the following: Given an $\epsilon>0$ there is a $\delta=\delta(\epsilon)>0$ such that $$0\leq r:=\sqrt{x^2+y^2}<\delta\qquad\Longrightarrow\qquad \bigl|f(x,y)\bigr|<\epsilon\ .$$
Of course it is allowed to use standard limit theorems, like the squeeze theorem, instead of $\epsilon$ and $\delta$. In the case at hand one has
$$|f(x,y)|={r^2\>|\cos\phi\sin\phi| \over r \sqrt{2\cos^2\phi+\sin^2\phi}}\leq r\ ,$$
and this implies $\lim_{(x,y)\to0} f(x,y)=0$.
A: You can use polar coordinates to evaluate your limit which becomes

$$\lim_{r\to 0} \frac{r^2\cos \theta \sin \theta}{r\ \sqrt{2\cos^2 \theta +\sin^2 \theta}}   = \lim_{r\to 0} \frac{r \cos \theta \sin \theta}{ \sqrt{2\cos^2 \theta +\sin^2 \theta}} =0 $$

