How does one take limit along a path? So in multivariable calculus for a limit of a function to exist, the limits of the function along all possible paths must exist and equal the same value.  
But how does one calculate the limit along a given path?
Say I have $f(x,y) = (x^3(y+1), x-y^2)$ (this is off the top of my head, so sorry if it doesn't work out well) and I want to know what the value of $\lim_{(x,y) \to (2,3)}$ is along the path $y= x^2 -1$.  How would I calculate the limit?

My guess is we'd need to parametrize the curve?  In that case, I guess we'd need some $\phi(t): [a,b] \subset \Bbb R \to \Bbb R^2$.  So we'll let $t=x$, then $\phi(t) = (t, t^2-1)$.  Then would $\lim_{(x,y) \to (2,3)} f(x,y) = \lim_{t \to 2} f(\phi(t))$.  I don't know for sure, bu it seems like we'd need something like the chain rule here -- that is can we just plug stuff in like this?

Moreover, is $\frac {\partial f}{\partial x}$ the limit of the gradient in the x-direction?  If so, is there some notation that tells us what the limit of the gradient at a point is along some other given path (like $y=x^2 -1$ for instance)?
I could be way off here.  I just realized that even though I know how to calculate partials and gradients and multiple integrals, I don't really understand the basics.
 A: *

*In your example, along the curve defined by $y=x^2-1$, the limit of $f(x,y) = (x^3(y+1), x-y^2)$ as $(x,y)\to (2,3)$ is 
$$
\lim_{x\to 2}f(x,x^2-1)=\lim_{x\to 2}(x^5,x-(x^2-1))=(32,-1).
$$
There is no chain rule here since we are not taking derivatives.

*In general, let $f:U\to \mathbb{R}^2$ be a map where $U$ is a non-empty open subset of $\mathbb{R}^2$. Let $p_0=(x_0,y_0)\in U$. Suppose $p_0$ is a limit point of $S\subset U.$ Then one can talk about the limit (if exists)
$$
\lim_{(x,y)\to p_0\\(x,y)\in S}f(x,y)
$$
where we consider the subspace topology on $S$ and the standard metric topology on $\mathbb{R}^2$. One such example is given by $S$ being a (smooth) curve in $U$ passing through the point $p_0$.

*Since your example of $f=(f_1,f_2)$ is a vector field, $\frac{\partial f}{\partial x}=(\frac{\partial f_1}{\partial x},\frac{\partial f_2}{\partial x})$. The partial derivatives:
$$
\frac{\partial f_i}{\partial x}(x,y)=\lim_{t\to 0}\frac{f_i(x+t,y)-f_i(x,y)}{t}.
$$

*"And is $\frac {\partial f}{\partial x}$ the limit of the gradient in the $x$-direction?"
Not even wrong. The gradient of $f$, given by
$$
(\frac{\partial f_1}{\partial x},\frac{\partial f_2}{\partial x}),
$$
is a limit. On the other hand, $\frac{\partial f_i}{\partial x}(x,y)$ is the directional derivative of $f_i$ at $(x,y)$ in the direction $(0,1)$, or as you put it, the $x$-direction.
