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?


2 Answers 2


Your hunch is correct. Letting $u = \frac{n}{r}$, we have: $$\begin{align*} \lim_{n\to\infty}\left(1 + \frac{r}{n}\right)^n &= \lim_{n\to\infty}\left(\left(1+\frac{r}{n}\right)^{n/r}\right)^r\\ &= \lim_{u\to\infty}\left(\left(1 + \frac{1}{u}\right)^u\right)^r\\ &= \left(\lim_{u\to\infty}\left(1 + \frac{1}{u}\right)^u\right)^r\\ &= e^r. \end{align*}$$

  • $\begingroup$ I will add this proof but its a bit late. +1. $\endgroup$
    – GarouDan
    Mar 6, 2012 at 20:06

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 Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.