If $x(z)$ and $y(z)$ are analytic with $x(0) = 0 = y(0)$ then $x(z)^{y(z)} \to 0$ as $z \to 0$ I'm a programmer and my math is a little rusty, but usually sufficient for my needs.
However, I came across the following statement in the exp(3) manual page of the libm C library that has me stumped:

Analysts will accept $0^0 = 1$ despite that $x^y$ can approach anything or nothing as $x$ and $y$ approach $0$ independently. The reason for setting 0**0 = 1 anyway is this:
  If $x(z)$ and $y(z)$ are any functions analytic (expandable in power series) in $z$ around $z = 0$, and if there $x(0) = y(0) = 0$, then $x(z)^{y(z)} \to 1$ as $z \to 0$.

The intention of the first sentence is perfectly clear.
I am having trouble making sense of the second statement. How (and where) is $x(z)^{y(z)}$ even defined?
The usual way of dealing with such an expression would be to write something like $\exp\big(\log(x(z)) \cdot y(z)\big)$, but as $x(0) = 0$, the logarithm $\log(x(z))$ is not defined in a neighborhood of $0$, so $z \to 0$ would need to be qualified.
Even the most basic case $z^z$ needs some form of interpretation since we're in trouble for negative $z$:

Question: Is there a way of interpreting $x(z)^{y(z)}$ in a pointed neighborhood of $0$ that doesn't involve branches of the logarithm and, if so, is the statement above correct with this interpretation? If not, what would be a correct interpretation of the above statement?

 A: One way to define $x(z)^{y(z)}$ is to use the universal cover of $\mathbb{C}^\times = \mathbb{C} \setminus \{0\} $, which can be defined by the map :
$$ \mathbb{C} \rightarrow \mathbb{C}^\times, t \mapsto z=\exp(t).$$
When $\mathrm{Re}(t) \to -\infty$ then $z \to 0$.
Hence the two functions $z \mapsto x(z)$ and $z \mapsto y(z)$ on $\mathbb{C}^\times$ define two functions on the universal cover $\mathbb{C} $ by $t \mapsto x(\exp(t))$ and $t \mapsto y(\exp(t))$ (at least locally).
Recall that for any homolophic function $f : \mathbb{C} \rightarrow \mathbb{C}^\times$, there exists an holomorphic function $g : \mathbb{C} \rightarrow \mathbb{C}$ such that $f = \exp \circ g$. Applying this to $f : t \mapsto x(\exp(t)) = x(z)$, we conclude that there exists a function $X$ such that
$$x(\exp(t)) = \exp(X(t)).$$
Hence we can define $t \mapsto x(z)^{y(z)} = x(\exp{t})^{y(\exp{t})}$, by
$$x(\exp{t})^{y(\exp{t})} = \exp \big( X(t).y(\exp(t)) \big).$$
And we have :

$$ \lim_{\mathrm{Re}(t) \to -\infty} x(\exp{t})^{y(\exp{t})} = 1.$$

Proof: write $x(z) = z^n(a+o(1))$ and $y(z) = z^m(b+ o(1))$.
Then we can write:
$$ x(\exp(t)) = \exp( nt + \ln a + o(1)) \text{ and } y(\exp(t)) = \exp(mt).(b+o(1)).$$
Hence
$$x(\exp{t})^{y(\exp{t})} = \exp[ (nt + \ln a + o(1)).\exp(mt).(b+o(1))].$$
This goes to $1$ when $\mathrm{Re}(t) \to -\infty$, because $nt.\exp(mt)$ goes to $0$.
