Derivative of Exponential at Zero

Theorem
Let $$\exp x$$ be the exponent of $$x$$ for real $$x$$.

Then $$\lim_{x \to 0} \frac {\exp x - 1} {x} = 1$$.

Proof
L'Hôpital's rule gives $$\lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$$ (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 $$\lim_{x \to 0} \frac {\exp x - 1} {x} = \lim_{x \to 0} \frac {\exp x} {1} = \exp 0 = 1$$.