$\exp(x)$ 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.
 A: 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.
