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 &amp;= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\
&amp;= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\
&amp;= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\
&amp;= e^r.
\end{align}

This argument is fine if $r &gt; 0$ since $u \rightarrow \infty$ as $n \rightarrow \infty$, but when $r &lt; 0$, $u \rightarrow - \infty$ 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 $r &lt; 0$:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &amp;= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\
&amp;= \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.
\begin{align*}\lim_{n\to\infty}\left( 1 - \frac{1}{n} \right)^n &amp;= \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} \\ &amp;= \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) \\ 
&amp;= \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} \\
&amp;= e^{-1}
\end{align*}
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 $\exp(r)$.