# Derivative of Exponential at Zero

From ProofWiki

## Theorem

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

Then:

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

## Proof 1

For all $x \in \R$, we have the following:

- $\exp 0 - 1 = 0$ from Exponential of Zero

- $D_x \left({\exp x - 1}\right) = \exp x$ from Sum Rule for Derivatives

- $D_x x = 1$ from Derivative of Identity Function.

Having verified its prerequisites, Corollary 1 to L'Hôpital's Rule yields immediately:

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

$\blacksquare$

## Proof 2

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

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

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac{\exp h - 1} h\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac{\lim_{n \to \infty} \left({1 + \dfrac h n}\right)^n - 1} h\) | \(\displaystyle \) | \(\displaystyle \) | by definition of the exponential | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac{\displaystyle \lim_{n \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\) | \(\displaystyle \) | \(\displaystyle \) | Binomial Theorem | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \lim_{n \to \infty} \frac{\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\) | \(\displaystyle \) | \(\displaystyle \) | as $h$ is constant | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \lim_{n \to \infty} \left({ {n \choose 0} 1 - 1 + {n \choose 1} \left({\frac h n}\right)\frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \left({\frac h n}\right)^k \frac 1 h }\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \lim_{n \to \infty}1 + \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k-1} }{n^k}\) | \(\displaystyle \) | \(\displaystyle \) | Powers of Group Elements | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1 + h \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac{h^{k-2} } {n^k}\) | \(\displaystyle \) | \(\displaystyle \) | Powers of Group Elements |

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

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

$\blacksquare$

## Proof 3

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {e^x - 1} x\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac {e^x - e^0} x\) | \(\displaystyle \) | \(\displaystyle \) | Exponential of Zero | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\to\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left.{\dfrac {\mathrm d} {\mathrm dx} e^x}\right \vert_{x \mathop = 0}\) | \(\displaystyle \) | \(\displaystyle \) | as $x \to 0$, from definition of derivative at a point | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle e^x \vert_{x \mathop = 0}\) | \(\displaystyle \) | \(\displaystyle \) | Derivative of Exponential Function | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | Exponential of Zero |

$\blacksquare$