(related to Proposition: \(\exp(0)=1\))
According to the estimate for the remainder term of the exponential function, we have \[\exp(x)=\sum_{n=0}^N\frac{x^n}{n! }+r_{N + 1}(x),\] \[|r_{N + 1}(x)|\le 2\frac{|x|^{N+1}}{(N+1)!}\quad\text{for }|x|\le \frac {N+2}2.\] It follows for \(N=0\) that
\[|\exp(x)-1|\le 2|x|\quad\text{for }|x|\le 1.\]
For any convergent real sequence \((x_n)_{n\in\mathbb N}\) with \(\lim_{n\to\infty} x_n=0\) it follows
\[\lim_{n\to\infty}|\exp(x_n)-1|=0.\] Thus \[\lim_{n\to\infty}\exp(x_n)=1.\]
