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
L'Hôpital's rule gives:
 * $\displaystyle \lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$

(provided the appropriate conditions are fulfilled).

Here we have:
 * $\exp 0 - 1 = 0$


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


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

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