Derivative of Exponential at Zero/Proof 1
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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): $\S 14.5 \ (3) \ \text{(i)}$