Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$
Then, textbooks usually derive equation (1) in the following manner:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\
&= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\
&= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\
&= e^r.
\end{align}
This argument is fine if $r > 0$ since $u \rightarrow \infty$ as $n \rightarrow \infty$, but when $r < 0$, $u \rightarrow - \infty$ as $n \rightarrow \infty$.
How can I extend the proof for (1) where $r$ is any real number?
When $r = 0$, $\lim_{n \rightarrow \infty} (1+0/n)^n = 1$ (Although, I should be careful about evaluating limits that look like $``1^{\infty}"$.)
Here's my attempt so far for the case where $r < 0$:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\
&= \left(\lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^u\right)^{-r}.
\end{align}
My question boils down to how to show the following limit from the definition above for $e$ $$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.$$
Thanks.
 A: Consider proving the reciprocal, i.e.
$$
\lim \left(1 - \frac 1n\right)^{-n} = \mathrm e. 
$$
The expression inside could be rewritten as 
$$
\left(1 - \frac 1n \right)^{-n} = \left(\frac {n-1}n\right)^{-n} = \left(\frac n{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \left(1 + \frac 1{n-1}\right). 
$$
Now use the definition of $\mathrm e$:
$$
\lim_n \left(1 + \frac 1{n-1}\right)^{n-1} = \mathrm e.
$$
Since 
$$
\lim_n \left(1 + \frac 1{n-1}\right) = 1, 
$$
we conclude that 
$$
\lim_n \left(1 - \frac 1n \right)^{-n} = \lim_n \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \lim_n \left(1 + \frac 1{n-1}\right) = \mathrm e \cdot 1 = \mathrm e. 
$$
by the law of arithmetic operations of limits. 
Therefore the original limit is $1/\mathrm  e$.
A: \begin{align*}\lim_{n\to\infty}\left( 1 - \frac{1}{n} \right)^n &= \lim_{n\to\infty}\left(\frac{n-1}{n}\right)^n = \lim_{n\to\infty}\left(\frac{1}{\frac{n}{n-1}}\right)^n = \lim_{n\to\infty}\frac{1}{\left(\frac{n}{n-1}\right)^n} \\ &= \lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n-1}\right)^n} = \lim_{n\to\infty} \left(\frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} \cdot \frac{1}{1 + \frac{1}{n-1}} \right) \\ 
&= \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} = \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n}\right)^{n}} = \frac{1}{e} \\
&= e^{-1}
\end{align*}
A: Let $P= \left (1 - \frac 1 n \right )^n.$ Then $\ln P = \frac {\ln \left (1 - \frac 1 n \right )} {\frac 1 n}.\ (*)$ 
Now as $n \rightarrow \infty$ then $(*)$ is a $\frac 0 0$ form. Applying L'Hospital we have $$\begin{align*} \lim\limits_{n \rightarrow \infty} \ln P & = \lim\limits_{n \rightarrow \infty} \frac {\frac {1} {\left (1 - \frac 1 n \right )} \cdot \frac {1} {n^2}} {-\frac {1} {n^2}}. \\ & = - \lim\limits_{n \rightarrow \infty} \frac {1} {\left (1 - \frac 1 n \right )}. \\ & = -1. \end{align*}$$
This shows that $$\lim\limits_{n \rightarrow \infty} P =\lim\limits_{n \rightarrow \infty}   \left (1 - \frac 1 n \right )^n = \frac 1 e.$$
A: The easiest way, working for all $r \in \mathbb{R}$, is to write
$$\left( 1+ \frac{r}{n}\right)^n =\exp \left( n \ln \left( 1+ \frac{r}{n}\right)\right) = \exp \left( n \left( \frac{r}{n} + o\left( \frac{1}{n}\right)  \right)\right) = \exp \left( r + o\left( 1\right)  \right)$$
So it converges to $\exp(r)$.
A: $f(x):=\log x;$
$f'(1)= \lim_{n \rightarrow \infty}\dfrac{f(1-1/n)-f(1)}{-(1/n)}$;
Hence
$\lim_{n \rightarrow \infty} (n\log(1-1/n)-\log 1)=$
$\lim_{n \rightarrow \infty} (-\dfrac{\log (1-1/n)-\log 1}{-(1/n)})=- \log '(1)=-1$
Then 
$\lim_{n \rightarrow \infty} \exp n(\log (1-1/n))=$
$\exp (\lim_{n \rightarrow \infty} n\log (1-1/n))=$
$\exp (-1)$.
Used: Continuity of $\exp$.
