# Derivative of Exponential at Zero

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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 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 | ||||||||||

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

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

$\blacksquare$