$\exp(x)$ as defined by a net Motivation:
So, I had an idle thought last week, and I thought I would ask it here before I forget about it.  It is well known that we can define
$$
e^x = \lim_{n \to \infty} \left(1 + \frac x{n} \right)^{n}
$$
Where $x$ here can be taken as either a number or a linear operator.
This is often intuitively explained as stating that $e^x$ is the multiplication which is generated by the infinitessimal perturbation of $1$ in the direction of $x$.  Or, if you prefer, $e^x$ is the "continuous interest rate" that is generated by the periodic interest rate $x$.
In either case, it is "suspicious" that we've broken the $n$th product down into $n$ identical pieces, so perhaps we can come up with a more "robust" definition.  In that vein:

Problem Statement
Let $\lambda$ denote a tuple $(\lambda_1,\lambda_2,\dots,\lambda_n)$.  Let $\Lambda$ denote the set of all such (finite) tuples of positive $\lambda_i$ satisfying $\sum_i\lambda_i = 1$ with the partial order
$$
(\lambda_{1,1},\dots,\lambda_{1,k_1},\lambda_{2,1},\dots,\lambda_{2,k_2}, \dots \dots  \dots,\lambda_{n,1},\dots,\lambda_{n,k_n}) \succeq\\
([\lambda_{1,1}+\cdots+\lambda_{1,k_1}],[\lambda_{2,1}+\cdots+\lambda_{2,k_2}],   \dots,[\lambda_{n,1}+\cdots+\lambda_{n,k_n}])
$$
So, for example, $(1/2,1/2) \preceq (1/4,1/4,1/2) \preceq (1/8,1/8,1/4,1/2)$.
Define the net $(e_{\lambda}^x)_{\lambda \in \Lambda}$ by
$$
e_{\lambda}^x = \prod_{i =1}^n \left( 1 + \lambda_i x\right)
$$

Conjecture: $\lim_{\lambda \in \Lambda} e_\lambda^x = e^x$

Is this statement correct?  Has this been done before?  Is this demonstrably useless?  Let me know.

Having come across one of my old questions, I've decided to leave an answer (following Daniel's hint in the comment).
Fix $x \in \Bbb C$. The Taylor expansion for $\log(1 + z)$ (with the Peano form of the remainder) tells us that
$$
\log(1 + \mu x) - \mu x = [1 + \zeta(\mu x)]\frac {\mu^2 x^2}2
$$
where $\zeta(z) \to 0$ as $z \to 0$.  For any $\epsilon > 0$, we may select a sufficiently small $\delta  > 0$ with $|\delta x| < \sqrt{\epsilon}$ such that $|\log(1 + \mu x) - \mu x| < \mu^2 |x|^2$ whenever $0 < \mu < \delta$.  If $\lambda = (\lambda_1,\dots,\lambda_n)$ is chosen such that $\max_k(\lambda_k) < \delta$, then we have
$$
|x - \log e_{\lambda}^x| = \sum_{k=1}^n [\lambda_i x - \log(1 + \lambda_i x)] \leq \sum_{k=1}^n\lambda_i^2 x^2 < \delta^2 x^2 < \epsilon
$$
Now, if we select $N$ with $1/N < \delta$, then the above shows that
$$
\lambda \succeq (1/N, \dots, 1/N) \implies |x - \log e^x_{\lambda}| < \epsilon
$$
Thus, we have shown that the net $\log e^x_{\lambda}$ converges to $x$, which means that the net $e_{\lambda}^x$ coverges to $e^x$, which was the desired conclusion.
