Are we allowed to "flatten" domains in calculus 3? If we're taking the limit of some function say $f(x,y)=xy^{xy}$ as $(x,y) \to (a,b)$ are we allowed to make the substitution $u = xy$ and essentially flatten the domain from 2D to 1D?
I.e., is it valid to do something like this:
$\lim_{(x,y)\to(0,b)}(xy)^{xy} = \lim_{u \to 0}u^u$
I'm not sure we are allowed to do this, because we've gone from infinite paths of $x,y$ to only two possible paths of $u$ (because it is now 1D)
 A: Yes. Essentially, what you are noticing is the following. We may write
$$g(u)=u^u$$
$$u(x,y)=xy$$
and then
$$f(x,y)=xy^{xy}=g(u(x,y))$$
this is the composition of continuous functions and we may write
$$\lim_{(x,y)\rightarrow (0,b)}g(u(x,y))$$
however, along any path, the inner function has a limit of $u(0,b)$. Thus, we have:
$$\lim_{u\rightarrow u(0,b)}g(u)$$
which is what you have. That is, we are using the fact that $(x,y)\mapsto xy$ is a continuous map to show that it respects all the structure necessary to make such a substitution.
A: Short answer: no.
Long answer:
The statement $\lim_{(x,y)\to (0,b)}$ implies that the limit exists and takes a unique value independent of the path chosen.
By setting $u=xy$, you are choosing a smaller family of paths. So the limit might exist in the smaller family, but not in the larger family.
A: In this case, the answer is yes.
Because
$$
\begin{align}
\left|\,xy\,\right|
&\le\left|\,x(y-b)\,\right|+\left|b\right|\left|\,x\vphantom{b}\,\right|\\
&\le\frac12\left(x^2+(y-b)^2\right)+\left|b\right|\left(x^2+(y-b)^2\right)^{1/2}\tag{1}
\end{align}
$$
if $(x,y)$ is close to $(0,b)$, then $xy$ is close to $0$. Therefore,
$$
\lim_{u\to0}g(u)=L\implies\lim_{(x,y)\to(0,b)}g(xy)=L\tag{2}
$$
In the question, $f(x,y)=g(xy)$ where $g(u)=u^u$. Applying $(2)$, we get
$$
\lim_{u\to0}g(0)=1\implies\lim_{(x,y)\to(0,b)}f(x,y)=\lim_{(x,y)\to(0,b)}g(xy)=1\tag{3}
$$
