Derivative of Exponential at Zero/Proof 2

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Theorem

Let $\exp x$ be the exponential of $x$ for real $x$.


Then:

$\displaystyle \lim_{x \mathop \to 0} \frac {\exp x - 1} x = 1$


Proof

Note that this proof does not presuppose Derivative of Exponential Function.

We use the definition of the exponential as a limit of a sequence:

\(\displaystyle \frac {\exp h - 1} h\) \(=\) \(\displaystyle \frac {\lim_{n \mathop \to \infty} \left({1 + \dfrac h n}\right)^n - 1} h\) Definition of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {\displaystyle \lim_{n \mathop \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\) Binomial Theorem
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \frac{\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\) as $h$ is constant
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \left({ {n \choose 0} 1 - 1 + {n \choose 1} \left({\frac h n}\right) \frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \left({\frac h n}\right)^k \frac 1 h }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty}1 + \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 1} }{n^k}\) Powers of Group Elements
\(\displaystyle \) \(=\) \(\displaystyle 1 + h \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 2} } {n^k}\) Powers of Group Elements

The right summand converges to zero as $h \to 0$, and so:

$\displaystyle \lim_{h \mathop \to 0} \frac {\exp h - 1} h = 1$

$\blacksquare$