Can limits to (positive) infinity be replaced by ordinals?

Question

For example, does the following hold?

$$\left(1+\frac1\omega\right)^\omega=e$$

Answer 1

The expression $\left(1+\frac{1}{\omega}\right)^{\omega}$ doesn't make any sense, since there is no division by ordinals or raising a real to ordinal power. However, there is a way this can be made a formal statement.

A sequence of elements of $X$ is just a function $f: \omega \rightarrow X$. Now, consider $\omega +1$ with order topology: here, the point $\omega$ is the limit of the sequence $\{0, 1, 2, \dots\}$ of all natural numbers. So, a sequence $f:\omega\rightarrow X$ has a limit iff the function $f$ can be extended to a continuous function $g:\omega+1\rightarrow X$: the limit will be $g(\omega)$.

Now, consider a function $f: \omega \rightarrow \mathbb{R}$: $f(n)=\left(1+\frac{1}{n}\right)^n$. As we know, $\lim\limits_{n\rightarrow\infty}f(n)=e$. So, this function can be extended to a function $g:\omega+1\rightarrow\mathbb{R}$ such that $g(n)=f(n)$ for $n \in \omega$ and $g(\omega)=e$.

Answer 2

No. $(1 + \frac{1}{x})^x $ represents an intangible expression. $x$ can take any value and by calculating it, it will not give you $e$. It's the series from $0$ to $\infty $ or the $\lim{x \to \infty}$ that gives you $e$.