How to proof that $\lim_{h \to 0}\frac{e^h-1}{h} = 1$ using the definition $e = \lim_{n \to \infty}(1+\frac{1}{n})^n$?

Question

In other words, how I can prove that these two definitions of $e$ is equal? I saw these two definitions while trying to find proofs for $\frac{d}{dx}e^x$ and $\frac{d}{dx}\ln x$; some use the former definition, and others used the latter, and I cannot find a proof that these two definitions are equal. So how do I prove this? Thank you!

Answer 1

From $e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n$ easily follows $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$ for $x>0$.

If you prove that even $e=\lim\limits_{n\to-\infty}\left(1+\frac1n\right)^n$, then $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$ also for $x<0$.

Expand the expression inside the limit: $$\left(1+\frac xn\right)^n=\sum_{k=0}^n\binom nk\frac{x^k}{n^k}=\sum_{k=0}^n\frac{n(n-1)\ldots(n-k+1)}{n^k}\frac{x^k}{k!}$$

For a fixed $k$ we have $\frac{n(n-1)\ldots(n-k+1)}{n^k}\to1$, so it's not hard to see $$\left(1+\frac xn\right)^n\to\sum\limits_{k=0}^\infty\frac{x^k}{k!}$$ using the fact that $\sum\limits_{k=0}^\infty\frac{x^k}{k!}<\infty$ and that $\frac{n(n-1)\ldots(n-k+1)}{n^k}<1$.

Your limit follows easily from this (in fact, we found the whole Taylor series for $e^x$).