Derivative of Exponential at Zero

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 and One


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


 * $D_x x = 1$ from Differentiation of the Identity Function.

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


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

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:

The right summand converges to zero as $h \to 0$, and so:
 * $\displaystyle \lim_{h \to 0}\frac{\exp h - 1} h = 1$