The preamble of my analysis book defines $\exp$ in terms of its Taylor series around $0$.
$$ \exp(z) \stackrel{def}{=} \sum_{n=0}^\infty \frac{z^n}{n!} \tag{1} $$
I'm trying to show that this is equal to $e^z$, using the more familiar limit definition of $e$ .
$$ e \stackrel{def}{=} \lim_{n \rightarrow \infty} \left( 1+\frac{1}{n}\right) ^n \tag{2} $$
For one of steps in the proof I want to invoke the binomial theorem and translate $(3)$ to $(4)$
$$ \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n} \right)^{n} \tag{3} $$
to
$$ \lim_{n \in \mathbb{N} \rightarrow \infty} \sum_{i=0}^n {n \choose i} \frac{x^i}{n^i} \tag{4} $$
I've changed $n$ from ranging over real numbers $\mathbb{R}$ to ranging over natural numbers $\mathbb{N}$.
In this particular case, I think the step "works" because $(1+ \frac{x}{n})^n$ is fairly well behaved with respect to $n$.
In general, however, I could have a pathological function that's defined differently on $\mathbb{N} + 0.5$, for example. In which case the original expression would not converge, but the rewritten one with the "sparse" limit would converge.
Is there a name in general for such "sparse limits" and criteria for when you're allowed to use them?