Derivative of Exponential at Zero/Proof 1

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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

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:

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

$\blacksquare$


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