A Limit that Converges to e

$\huge\boxed{\lim_{n\to\infty}(1+\frac 1n)^n=e}$ (L’Hopital’s Rule Proof)

This limit is a useful and interesting result to know. Note especially that the method “$\lim_{n\to\infty}(1+\frac 1n)^n=1^\infty=1$” is incorrect.

Proof:

We will prove $\lim_{x\to\infty}(1+\frac 1x)^x=e$ instead, and this implies $\displaystyle \lim_{n\to\infty}(1+\frac 1n)^n=e.$

First, we will find the limit $\lim_{x\to\infty}\ln(1+\frac 1x)^x$.

\begin{aligned} \lim_{x\to\infty}\ln(1+\frac 1x)^x&=\lim_{x\to\infty}x\ln(1+\frac 1x)\ \ \ \text{(Bringing down the power)}\\ &=\lim_{x\to\infty}\frac{\ln (1+x^{-1})}{x^{-1}}\\ &=\lim_{x\to\infty}\frac{\frac{1}{1+x^{-1}}(-x^{-2})}{-x^{-2}}\ \ \ \text{(L'Hopital's Rule)}\\ &=\lim_{x\to\infty}\frac{1}{1+x^{-1}}\\ &=1. \end{aligned}

So $\lim_{x\to\infty}(1+\frac 1x)^x=e^1$.

Exercise

If you are interested, you can try to prove $\displaystyle \lim_{n\to\infty}(1+\frac cn)^n=e^c$ where $c\in\mathbb{R}$.