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.

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.
