# $\lim_{n\rightarrow \infty}(1+\frac{r}{n})^n$ 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?

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Note that this leads to continuously compounding interest‌​. – Mark Hurd Mar 3 '12 at 13:36
Try L'Hospital's Rule. – Patrick Mar 3 '12 at 14:16

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, $L=e^r$.
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*}