$$ % Define your custom commands here \newcommand{\bmat}[1]{\begin{bmatrix}#1\end{bmatrix}} \newcommand{\E}{\mathbb{E}} \newcommand{\P}{\mathbb{P}} \newcommand{\S}{\mathbb{S}} \newcommand{\R}{\mathbb{R}} \newcommand{\S}{\mathbb{S}} \newcommand{\norm}[2]{\|{#1}\|_{{}_{#2}}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\pdd}[2]{\frac{\partial^2 #1}{\partial #2^2}} \newcommand{\vectornorm}[1]{\left|\left|#1\right|\right|} \newcommand{\abs}[1]{\left|{#1}\right|} \newcommand{\mbf}[1]{\mathbf{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\bm}[1]{\boldsymbol{#1}} \newcommand{\nicefrac}[2]{{}^{#1}\!/_{\!#2}} \newcommand{\argmin}{\operatorname*{arg\,min}} \newcommand{\argmax}{\operatorname*{arg\,max}} \newcommand{\dd}{\operatorname{d}\!} $$

Famous limit

calculus
An attractive way to compute this famous limit.
Author

Aykut C. Satici

Published

September 25, 2026

Problem 1 (Famous limit) Show that \[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e. \]

Answer 1 Let \(a_n = \left(1 + \frac{1}{n}\right)^n\). Then we can write \[ \ln(a_n) = n \ln\left(1 + \frac{1}{n}\right) = \frac{\ln \left(1 + \frac{1}{n}\right) - \ln 1}{\frac{1}{n}}. \] Taking the limit as \(n \to \infty\) and using the definition of the derivative, we obtain \[ \lim_{n \to \infty} \ln(a_n) = \ln^\prime(1) = 1. \] Since \(\exp: \mathbb{R} \to \mathbb{R}\) is a continuous function, exponentiating both sides gives \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \exp(\ln(a_n)) = \exp\left(\lim_{n \to \infty} \ln(a_n)\right) = \exp(1) = e. \tag*{$\square$} \]

← Back to Math