Derivative of Exponential at Zero/Proof 1

Theorem

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

Then:

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

Proof

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:

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

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 14.5 \ (3) \ \text{(i)}$