$\lim_{n\rightarrow \infty}(1+\frac{r}{n})^n$ is equal to ${e^{r}}$? Since 
$$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n={e}$$
My strong hunch is that the following statement must also be true
$$\lim_{n\rightarrow \infty}\left(1+\frac{r}{n}\right)^n = {e^{r}}$$
for all $r>0$.
But I can neither prove or disprove it, any idea on how to prove it? Or if the statement is not true, how it should be modified so that it is true?
 A: Your hunch is correct. Letting $u = \frac{n}{r}$, we have:
$$\begin{align*}
\lim_{n\to\infty}\left(1 + \frac{r}{n}\right)^n &= \lim_{n\to\infty}\left(\left(1+\frac{r}{n}\right)^{n/r}\right)^r\\
&= \lim_{u\to\infty}\left(\left(1 + \frac{1}{u}\right)^u\right)^r\\
&= \left(\lim_{u\to\infty}\left(1 + \frac{1}{u}\right)^u\right)^r\\
&= e^r.
\end{align*}$$
A: Another way to see this. Suppose 
$$\lim_{n\to \infty} \left(1+\frac{r}{n}\right)^n = L.$$
Let us calculate $\ln(L)$:
$$\begin{align*}
\ln(L) &= \ln\left(\lim_{n\to \infty} \left(1+\frac{r}{n}\right)^n \right)\\
&=\lim_{n\to \infty} \ln\left(\left(1+\frac{r}{n}\right)^n\right)\\
&=\lim_{n\to \infty} n\ln\left(1+\frac{r}{n} \right)\\
&=\lim_{n\to \infty} \frac{\ln\left(1+\frac{r}{n} \right)}{\frac{1}{n}}\\
&=\lim_{n\to\infty} \frac{\frac{1}{1+\frac{r}{n}}\cdot\frac{-r}{n^2}}{-\frac{1}{n^2}}\\
&=\lim_{n\to\infty} \frac{r}{1+\frac{r}{n}}\\
&=r,
\end{align*}$$
where we have used the fact that $\ln(x)$ is continuous in $(0,\infty)$, and l'Hôpital's rule. Thus, $\ln(L)=r$, or equivalently, $L=e^r$.
