Showing that $\lim_{n \rightarrow \infty} \left( 1 + \frac{r}{n} \right)^{n} = e^{r} $. [duplicate]

$$ \lim_{n\to\infty}\left(1+\frac rn\right)^n=\lim_{n\to\infty}\left(\left(1+\frac rn\right)^{n/r}\right)^r =\left(\lim_{n/r\to\infty}\left(1+\frac rn\right)^{n/r}\right)^r=e^r $$

This is my take on the issue. I'm assuming the existence of all the limits involved simply because I just don't want to type that much more.

I am not using the exponential function to define exponents instead I am taking the definitions I used in this answer. Basically positive integer powers are repeated multiplication, rational powers through $n'th$ roots, real powers by least upper bounds. Negative powers are taken to be the reciprocals of the positive powers.

I may have made some typos or errors in judgement since the answer is quite long. I am happy to fix any issues which are brought to my attention.

We start with $e$ being defined as the limit as $n\rightarrow \infty$ of $(1+1/n)^n$. The goal is to establish that $(1+r/n)^n \rightarrow e^r$ for rational $r$ and to then use this result to define real powers of $e$. As a starting point we consider integer powers of $e$ first.

Positive Integer Powers Suppose that $k$ is a positive integer and consider the $k$'th power of $e$ defined as the product of $e$ with itself $k$ times.

$$ e^k = \prod_{i=1}^k \left[ \lim_{n\rightarrow \infty}(1+1/n)^n \right] = \lim_{n\rightarrow \infty} \prod_{i=1}^k \left[ (1+1/n)^n \right] = \lim_{n\rightarrow \infty} \left[ \prod_{i=1}^k (1+1/n) \right]^n $$

Note that, $$\prod_{i=1}^k (1+1/n) = (1+1/n)^k = \sum_{p=0}^k {k\choose p}\frac{1}{n^p} = 1+\frac{k}{n}+O(1/n^2) = (1+k/n)\left[1+O\left(\frac{1}{n^2(1+k/n)}\right)\right] $$

Note that $ 1/n^2 \geq 1/[n^2(1+k/n)]$ which means we can replace the $O(1/[n^2(1+k/n)])$ with $O(1/n^2)$. So that we have,

$$(1+1/n)^k = (1+k/n)\left[1+O\left(1/n^2\right)\right]$$

Taking the $n$'th power of this expression we get,

$$ (1+1/n)^{nk} = (1+k/n)^n\left[1+O\left(1/n^2\right)\right]^n = (1+k/n)^n \left[1+O\left(1/n\right)\right] $$

$$ \Rightarrow \vert (1+1/n)^{nk} - (1+k/n)^n \vert = O(1/n) $$

This tells us that the limit of $(1+1/n)^{nk}$ is the same as the limit of $(1+k/n)^n$ as $n\rightarrow \infty$. Therefore we conclude that,

$$\boxed{ e^k = \lim_{n\rightarrow \infty} (1+1/n)^{nk} = \lim_{n\rightarrow \infty} (1+k/n)^n} $$

Positive Rational Powers

To include positive rational powers we just need to establish that $e^{p/q}$ is the $q$'th root of $e^p$. We suspect that $(1+p/qn)^n\rightarrow e^{p/q}$ as $n\rightarrow \infty$ which motivates us to take the $q$'th power of this expression in order to compare it with $e^p$.

$$ (1+p/nq)^{nq} = \left( 1+ p/n + \sum_{i=2}^q {q\choose i } \frac{p^i}{(nq)^i}\right)^n = \left(1+p/n + O\left( 1/n^2 \right)\right)^n = (1+p/n)^n\left(1+O\left(1/n^2\right)\right)^n = (1+p/n)^n \left(1+O\left(1/n\right) \right)$$

We can then use this to show that $ \vert (1+p/nq)^{nq} - (1+p/n)^n \vert = O(1/n)$ which by the same argument as above tells us that the two expressions have the same limit.

So we now have that the limit of $(1+p/nq)^n$, which we will call $L$, has the property that $L^q=e^p$. Therefore $L$ is the $q$'th root of $e^p$ and we indicate this with the notation $e^{p/q}$.

$$ \boxed{(1+p/nq)^n \rightarrow e^{p/q}} $$ ** Negative Rational Powers **

Negative rational powers of a number are defined as the reciprocals of the positive rational powers of the same number. Therefore given a positive rational number $r$ we wish to show that $(1-r/n)^n\rightarrow 1/e^r$.

$$ \left( \lim_{n\rightarrow \infty} (1-r/n)^n \right) \left( \lim_{n\rightarrow \infty} (1+r/n)^n \right) = \lim_{n\rightarrow \infty} \left[ (1-r/n) (1+r/n)\right]^n = \lim_{n\rightarrow \infty} \left[ 1-r^2/n^2\right]^n = \lim_{n\rightarrow \infty} \left[ 1-O(1/n) \right] = 1 $$

Therefore the limit of $(1-r/n)^n$ is the reciprocal of $e^r$ and we identify it with the notation $e^{-r}$.

$$ \boxed{(1-r/n)^n \rightarrow e^{-r}}$$

** Real / Irrational Powers of e **

We now understand that for rational powers $e^r= \lim_{n\rightarrow \infty} (1+r/n)^n$. We wish to show that this still holds for irrational powers $x$.

The definition of a real power $e^x$ is in terms of the least upper bound property of the real numbers,

$$ e^x = \sup \lbrace e^r \mid r\in Q, r < x \rbrace $$

Our job is then to show that the limit of $(1+x/n)^n = L$ is the least upper bound of the set indicated above.

Note that given a rational number $r$ which is less than $x$ we have, $(1+r/n) < (1+x/n)$ which tells us that $(1+r/n)^n < (1+x/n)^n$. Allowing $n\rightarrow \infty$ we get the inequality $e^r \leq L $. We have therefore established that $L$ is an upperbound on the indicated set of rational powers of $e$.

To show that $L$ is the least upper bound we must show that no number less than $L$ can be an upper bound of the set. We will establish this by showing that $e^r$ can be made arbitrarily close to $L$ by choosing an $r$ close to $x$.

Suppose that $r<x$ and $\vert x-r\vert < \delta $ then we have,

$$ \vert (1+x/n)^n - (1+r/n)^n \vert = \vert 1+x/n-1-r/n \vert \vert \sum_{i=1}^{n} \left[(1+x/n)^{n-i}(1+r/n)^{i-1} \right] \vert $$

$$ \leq \frac{\vert x-r \vert}{n} \sum_{i=1}^{n} \vert[(1+x/n)^{n-1} \vert \leq \frac{\delta}{n} n \vert (1+x/n)^{n-1} \vert = \delta (1+x/n)^{n-1}$$

Allowing $n\rightarrow \infty$ we get,

$$\vert e^r - L \vert \leq \delta L$$

Which means that $e^r$ can be put within $\epsilon >0$ of $L$ if we choose $r$ to be within $\delta = \epsilon L$ of $x$. Since epsilon can be made arbitrarily small we know that for any $\alpha<L$ we can find a $\alpha < e^r <L$. That is to say that $L$ is a least upper bound or the set of rational powers $e^r$ with $r<x$. Since this is the definition of $e^x$ we have shown that,

$$ \boxed{ (1+x/n)^n \rightarrow e^x } $$