(related to Proposition: Continuity of Exponential Function)

Let \(a\in\mathbb R\). We have to show that the exponential function \(\exp:\mathbb R\to \mathbb R\) is continuous, formally \[\lim_{x\to a}\exp(x)=\exp(a).\] Let \((x_n)_{n\in\mathbb N}\) be any convergent real series with \(\lim_{n\to\infty} x_n=a\). We have then \(\lim(x_n-a)=0\). Together with the result \(\exp(0)=1\) it follows \[\lim_{n\to \infty}\exp(x_n-a)=1.\] Because of the "non-zero property of the exponential function":bookofproofs$1417 \(\exp(x)\neq 0\) for all \(x\in\mathbb R\), and because of the functional equation of the exponential function5 we can conclude that \[1=\lim_{n\to \infty}\exp(x_n-a)=\frac{\lim_{n\to \infty}\exp(x_n)}{\lim_{n\to \infty}\exp(a)}=\lim_{n\to \infty}\frac{\exp(x_n)}{\exp(a)}.\] \[\exp(a)=\lim_{n\to \infty}\exp(x_n).\] In the last step we have used the formula for the quotient of convergent real sequences.

Thank you to the contributors under CC BY-SA 4.0!




  1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983