In $\lim(x,y)\to(0,0)$ why can I change to $(x^2,x)$? When we have a multivariable function  and we want to see if the function is continuous at a point, normally the origin, we sometimes "change" $(x,y)\to(0,0)$ to expressions like $(x^2,x)\to(0,0)$ to make it work. 
For example, for the function:
$f(x)=\begin{cases}\frac{yx^2}{(x^4+y^2)}&  \text{if } (x,y)\neq (0,0)\\
0& \text{if } (x,y)=(0,0)\end{cases}$
In order to see the discontinuity we can consider $(x,xm)\to(0,0)$ and we get $0$. But if we change to $(x,x^2)\to(0,0)$, the limit becomes $1/2$.
Why can we choose those curves?
Also, for a different function, could I choose $(1/x,x)\to(0,0)$. I know the limit of $1/x$ as $x$ goes to zero does not exist, therefore I am unsure.
 A: Confronted with such a problem you have to make a decision, founded on your experience with similar problems: Shall I try to prove that the limit exists, or shall I try to prove that the limit does not exist?
If you conjecture that the limit $\lim_{{\bf z}\to{\bf 0}}f({\bf z})$ does not exist you can try to exhibit two curves $$\gamma:\quad t\mapsto{\bf z}(t)=\bigl(x(t),y(t)\bigr)\ne{\bf 0},\qquad \lim_{t\to0+}{\bf z}(t)={\bf 0}\ ,\tag{1}$$
for which the limit  $\lim_{t\to0+}f\bigl({\bf z}(t)\bigr)$ is different, or one such curve, for which this limit does not exist. The logic behind this procedure is as follows: If  $\lim_{{\bf z}\to{\bf 0}}f({\bf z})=\alpha$ for a certain $\alpha$ then by the "law of nested limits" one has $\lim_{t\to0+}f\bigl({\bf z}(t)\bigr)=\alpha$ for all curves $(1)$.
If you conjecture that the limit $\lim_{{\bf z}\to{\bf 0}}f({\bf z})$ exists then you have to provide a fulfledged $\epsilon/\delta$ proof of this conjecture, and you cannot resort to special curves for a proof. In such cases it often, but not always, helps to express $f$ in polar coordinates, because the variable $r$ encodes the nearness of ${\bf z}$ to ${\bf 0}$ in a particularly simple way.
A: I think I know how to make this clear. 
Consider for example the function of one variable $g(x)=\begin{cases}1&x>0\\-1&x<0\end{cases}$ 

How would you show that $\lim\limits_ {x\to 0}g(x)$ does not exist? Well, you say that $\underbrace{\lim\limits_{x\downarrow 0}g(x)=1}_{\text{the limit from the right}}$ and $\underbrace{\lim\limits_{x\uparrow 0}g(x)=-1}_{\text{the limit from the left}}$. If the limit exists these should be equal. 
Why does it make sense to consider  this?
If we say that $\lim\limits _{x\to a}g(x)=L$ then this means that $g(x)$ gets increasingly close to $L$, if $x$ approaches $a$. No matter how $x$ approaches $a$. The "no matter how" part is really important. It would make very little sense to say that $\lim g(x) =1$, if this is not true when we approach from the left. 
Now, in the one-variable case  it is enough to check only the left-hand limit and the right-hand limit. Why? Well because $x$ can only approach $a$ in two ways:


*

*in a straight line form the right

*in a straight line from the left.


So in this case no matter how $x$ approaches $a$ is the same as saying no matter if $x$ approaches $a$ from the left or right.

Now for functions of more than one variable (for instance: two), not much changes. If we say that $\lim\limits_{(x,y)\to (0,0)}f(x,y)=L$, we want this to be true no matter how $(x,y)$ approaches $(0,0)$. The big difference is that there are now more than two ways for $(x,y)$ to approach $(0,0)$, or any other point. In fact there are infinitely many ways.  
Beacause there are infinitely many ways for $(x,y)$ to approach a point, this will not help you to actually find a limit. However, it can be usefull to show that a limit does not exist. For suppose that we find  there are (at least) two different ways to approach a point that give different outcomes. In that case we could no longer say that $f(x,y)$ gets close to $L$, no matter how... We have in stead shown that it does matter, thus the limit can't exist. 

In your example you may choose to approach $(0,0)$ along the paths $(x,mx)$ and $(x,x^2)$, because these are paths along which we can actually approach $(0,0)$. But we may just as well have chosen any other way to approach $(0,0)$.  These particular paths are usefull, because they give you different outcomes, immediately implying that the limit does not exist.  
You cannot actually approach $(0,0)$ along the path $(1/x,x)$, so this is not right.
