$\lim\limits_{(x,y)\to (3,2)}x^2y^3-4y^2$ I have some problems solving this:
$$\underset{(x,y)\to (3,2)}{\lim}(x^2y^3-4y^2)$$
I know that if $x = 3$ then $$\underset{y\to 2}{\lim}(3^2y^3-4y^2)=\underset{y\to 2}{\lim}(9y^3-4y^2)=56$$
Similarly, when $y = 2$, $$\underset{x\to 3}{\lim}(x^22^3-4(2^2))=\underset{x\to 3}{\lim}(x^28-16)=56$$
But then I get confused - I let $y = mx$ and it becomes $$\underset{x\to 3}{\lim}(x^2(mx)^3-4(mx)^2)=9(27)m^3-4(3)m^2$$. This equation depends on a value of m, let $m = 1$, then $$\underset{x\to 3}{\lim}=9(27)-12$$ which is not equal to 56, therefore I conclude that the limit DNE.
But it seems like this function should be continuous at point (3,2) since it has no restrictions and therefore it should have the limit point. 
P.S.: I feel like the situation could be different if I had to check the limit at (0,0), then I could use $y=x^2$ or $y=x$, but this (3,2) bugs me since I don't understand how to check it at this point.
 A: if you want to let y=mx and look at the behavior of the limit as x and y approach in that way, you have to make sure that $y \to 2$ as $x\to 3$. So you problem comes when you set $m=1$, which will make $y \to 3$ as $x\to 3$.
if you set $m=2/3$, then $y \to 2$ as $x\to 3$, and 
$\lim_{x\to 3}(x^2(mx)^3-4(mx)^2)=(2/3)^3x^5-4(2x/3)^2= 72-8=56 $ 
Let me make this clear though, this only shows that you do get the same limit as you use the $y=2x/3$ substitution, but it doesn't show that the limit exist. To show that the limit exist you have to use the continuity argument.
A: One way you can do it is to substitute in small level sets of circles. You probably know the circle equation $a^2+b^2 = R^2$, now move your point to (0,0) and replace $R$ with a positive number $\epsilon$, simplify and the result should go to the limit value as $\epsilon \to 0$.
A: One way you can tell if the limit exists is if you approach the limit from several different paths. For example, if you were to go along the x-axis you would test to see if 
\begin{equation}
f(x,y) -> L_{1} for all x = 0
\end{equation}
Then you would test along the y-axis.
\begin{equation}
f(x,y) -> L_{2} for all y = 0
\end{equation}
If $L_1 = L_2$, then you know the limit must exist. In fact, there's a possibility that $L_1$ is your limit. 
The always fail-proof way to find any limit is to approximate it by plugging values, very close to the point in question, into your calculator.
In this case, however, I believe you can the squeeze theorem to find it. 
