$e$ as the limit of a sequence

One of the standard definitions of $e$ is as

$$\lim_{n\rightarrow\infty}\left(1 + \frac{1}{n}\right)^n$$

But in all cases I've seen this limit, it is proven as a limit of the sequence $\Big\{\big(1 + \frac{1}{n}\big)^n\Big\}$, which seems to cover the limit for only $n$ as an integer. Now my question is whether this sequence limit is equivalent to a normal limit for which $n$ can be any real number. I can think of cases which this isn't generally true;

$$\lim_{n\rightarrow\infty}\ \sin(n\pi)$$

comes readily to mind, for which the limit as a sequence is simply $0$ but as a general limit, it is undefined. The limit for $e$ is used exactly as if it were a normal limit, which leads me to believe it is equivalent. Are there conditions for which the limit of a sequence and the corresponding function are identical?

Solution 1:

If $x>0$ be a real number, then write $x=n+y_x$ where $0 \leq y_x \leq 1$.

It is easy then to show that

$$ \left(1+\frac{1}{n+1} \right)^n \leq \left(1+\frac{1}{x}\right)^x \leq \left(1+\frac{1}{n}\right)^{n+1} \,.$$

Using this, you can prove that if

$$e=\lim_n \left(1+\frac{1}{n}\right)^n$$ then $$\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x =e$$

Edit

To complete the answer, in general if $f(x)$ is monotonic, then for sure

$$\lim_n f(n)= \lim_x f(x) \,.$$

I don't recall if $\left(1+\frac{1}{x}\right)^x$ is monotonic (and I am too lazy to differentiate it), but what we showed above is that it is at least "close to being monotonic". What I mean by this, I showed that $$f(n) h(n) \leq f(x) \leq f(n+1)g(n+1) \forall x\in [n, n+1)$$ where $h(n)$ and $g(n)$ go to $1$. The you basically sqeeze it.