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.


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.


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}.

Author: mathtuition88


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