Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists the same as $\lim_{n \to \infty} (1+1/n)^n$ exists

My expanded question:

Is showing
$\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists
as $z$ goes through real values
the same as $\lim_{n \to \infty} (1+\frac{1}{n})^n$ exists
as $n$ goes through integer values?
If not,
how much additional work
is needed to make the two

I am asking this because
I had posted a question which stated
a proof that the limit exists
for integer values.
I just a few minutes ago
answered a question
which involved the limit
as the argument
took on real values
by linking to my earlier answer.
The answer is here:
What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?

It then occurred to me
that this is not true
until it is proved that
the limit through integer values
is the same as
the limit through real values.

The proof through integer values
showed that
is an increasing sequence
and that
is an decreasing sequence,
but says nothing about
what happens between the
integer values.

If it can be shown
is an increasing function
of $z$,
that would be enough,
but I do not know of
a proof that is as elementary
as the proof that
$\lim_{n \to \infty}(1+1/n)^n$
The usual proofs I have seen
involve the power series
This can be proved in
a number of ways,
including starting with
$\ln'(z) = 1/z$.

I realize that
I am meandering,
so I’ll leave this
at this point.

Solutions Collecting From Web of "Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists the same as $\lim_{n \to \infty} (1+1/n)^n$ exists"

Let’s denote $a_n=(1+\dfrac1n)^n$ for the members of the sequence.

Because $0<n\le x<n+1$, then
Consequently we get the upper bound
and similarly the lower bound
So $(1+\dfrac1x)^x$ is sandwiched between two consecutive entries of the sequence $(a_n)$ multiplied by terms $\to 1$.

To answer: Yes, the two limit problems are equivalent.

By the definition of limits.

$c = \lim_{z \to \infty} (1+\frac{1}{z})^z$ exists if and only if
$\forall (s_n)_{n\in\mathbb{N}}, \, \, \left(\lim_{n \to \infty}(s_n) = +\infty \right)\implies \left(s = \lim_{n \to \infty} (1+\frac{1}{s_n})^{s_n} \,\mbox{ exists and }s = c\right)$.

Let $0 < x_1 < x_2$. Then $0 < \frac{x_1}{x_2} < 1$, so by the Bernoulli’s inequality

$$\left( 1 + \frac{1}{x_1} \right)^{x_1} = \left( 1 + \frac{1}{x_1} \right)^{\tfrac{x_1}{x_2} \cdot \ x_2} < \left( 1 + \frac{x_1}{x_2} \cdot \frac{1}{x_1} \right)^{x_2} = \left( 1 + \frac{1}{x_2} \right)^{x_2}.$$

Hence the function $f(x) = \left(1+\frac{1}{x}\right)^x$ is increasing on $(0, \infty)$.

The fundamental difference in dealing with limits $\lim_{n \to \infty}(1 + (1/n))^{n}$ and $\lim_{z \to \infty}(1 + (1/z))^{z}$ is that the conception of the first limit is simpler. The function $f(n) = (1 + (1/n))^{n}$ gives rational values for positive integers $n$ and can be calculated via simple arithmetic.

When you deal with $g(z) = (1 + (1/z))^{z}$ for positive real number $z$, then things are bit complex. The concept of an irrational exponent needs some development. If one has a sound theory of irrational exponents (some approaches are in my blog posts) then the proof that this limit exists is simple (based on taking logs).

In short the limit dealing with $n$ requires just theorems about monotone bounded sequences whereas limit concerning $z$ needs in addition the theory of arbitrary real exponents.