# Derivative of Exponential at Zero

## 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 1

For all $x \in \R$:

$\exp 0 - 1 = 0$ from Exponential of Zero
$\map {D_x} {\exp x - 1} = \exp x$ from Sum Rule for Derivatives
$D_x x = 1$ from Derivative of Identity Function.

Its prerequisites having been verified, Corollary 1 to L'Hôpital's Rule yields immediately:

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

$\blacksquare$

## 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:

 $\displaystyle \frac {\exp h - 1} h$ $=$ $\displaystyle \frac {\lim_{n \mathop \to \infty} \paren {1 + \dfrac h n}^n - 1} h$ Definition of Exponential Function $\displaystyle$ $=$ $\displaystyle \frac \lim_{n \mathop \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1}$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \frac \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1}$ as $h$ is constant $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren { {n \choose 0} 1 - 1 + {n \choose 1} \paren {\frac h n} \frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \paren {\frac h n}^k \frac 1 h}$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty}1 + \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 1} }{n^k}$ 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:

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

$\blacksquare$

## Proof 3

 $\displaystyle \frac {e^x - 1} x$ $=$ $\displaystyle \frac {e^x - e^0} x$ Exponential of Zero $\displaystyle$ $\to$ $\displaystyle \intlimits {\dfrac \d {\d x} e^x} {x \mathop = 0} {}$ Definition of Derivative of Real Function at Point, as $x \to 0$ $\displaystyle$ $=$ $\displaystyle \bigintlimits {e^x} {x \mathop = 0} {}$ Derivative of Exponential Function $\displaystyle$ $=$ $\displaystyle 1$ Exponential of Zero

$\blacksquare$