I was wondering if it is possible to get a link to a rigorous proof that

\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x

**Answer**

I would like to cite here an awesome German mathematician, Konrad Königsberger. He writes in his book *Analysis I* as follows:

Fundamental lemma. For every sequence of complex numbers w_n with a limit w it is true that \lim_{n \to \infty} \Bigl(1 + \frac{w_n}{n}\Bigr)^n = \sum_{k=0}^\infty \frac{w^k}{k!}.Proof. For every \varepsilon > 0 and sufficiently large index K we have the following estimations: \sum_{k=K}^\infty \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3 \quad\mbox{and}\quad |w_n| \le |w|+1.Therefore if n \ge K then \left|\Bigl(1 + \frac{w_n}{n}\Big)^n - \exp w \right| \le \sum_{k=0}^{K-1} \left|{n \choose k}\frac{w_n^k}{n^k} - \frac{w^k}{k!}\right| + \sum_{k=K}^n{n\choose k} \frac{|w_n|^k}{n^k} + \sum_{k=K}^\infty \frac{|w|^k}{k!}. The third sum is smaller than \varepsilon / 3 based on our estimations. We can find an upper bound for the middle one using {n \choose k} \frac 1 {n^k} = \frac{1}{k!} \prod_{i = 1}^{k-1} \Bigl(1 - \frac i n \Bigr) \le \frac 1 {k!}. Combining this with |w_n| \le |w| + 1, \sum_{k=K}^n {n \choose k} \frac{|w_n|^k}{n^k} < \sum_{k=K}^n \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3 Finally, the first sum converges to 0 due to w_n \to w and {n \choose k} n^{-k} \to \frac 1 {k!}. We can choose N > K such that it's smaller than \varepsilon / 3 as soon as n > N.

Really brilliant.

**Attribution***Source : Link , Question Author : Mai09el , Answer Author : GNUSupporter 8964民主女神 地下教會*