<s>[INST] Show $\lim\left ( 1+ \frac{1}{n} \right )^n = e$ if $e$ is defined by $\int_1^e \frac{1}{x} dx = 1$ I have managed to construct the following bound for $e$, which is defined as the unique positive number such that $\int_1^e \frac{dx}x = 1$.
$$\left ( 1+\frac{1}{n} \right )^n \leq e \leq \left (\frac{n}{n-1} \right )^n$$
From here, there must surely be a way to deduce the well-known equality
$$\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = e$$
I have come up with the following, but I am not absolutely certain if this is correct or not.
PROPOSED SOLUTION:
The lower bound is fine as it is, so we shall leave it alone. Note that 
$$\begin{align*}
\left ( \frac{n}{n-1} \right )^n &= \left ( 1+\frac{1}{n-1} \right )^{n} \\ &= 
\left ( 1+\frac{1}{n-1} \right )^{n-1} \left ( 1+\frac{1}{n-1} \right )
\end{align*}$$
So using the fact that the limit distributes over multiplication, we have
$$\lim_{n \rightarrow \infty} \left ( \frac{n}{n-1} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right )^{n-1} \lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right ) $$
Since $$\lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right ) = 1 $$
and 
$$\lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right )^{n-1} = \lim_{m \rightarrow \infty} \left ( 1+\frac{1}{m} \right )^m = e $$
We then have the required result
$$\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = e$$
 [/INST] "$ ( 1+\frac{1}{n}  )^n \leq e \leq  (\frac{n}{n-1})^n$"$
I'm assuming you've shown $( 1+\frac{1}{n}  )^n < ( 1+\frac{1}{n+1}  )^{n+1}$ so we can say $\lim_{n\rightarrow \infty}( 1+\frac{1}{n}  )^n = c \le e$ and likewise $\lim_{n\rightarrow \infty}(\frac{n}{n-1}  )^n = d \ge e$.
So $\frac cd = \lim_{n\rightarrow \infty}\frac{( 1+\frac{1}{n}  )^n}{(\frac{n}{n-1}  )^n}= \lim_{n\rightarrow \infty}(\frac{(n+1)(n-1)}{n^2})^n=$
$\lim_{n\rightarrow \infty}(1- \frac{1}{n^2})^n = \lim 1 - \frac{n}{n^2} = 1$
So $c = d$ and $c = e = d$.
So $c = \lim ( 1+\frac{1}{n}  )^n = e$.
</s>