Derivative of Exponential at Zero/Proof 4
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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
As we consider $x \to 0$, we may assume that $0 < \size x \le 1$.
Then:
\(\ds \size {\frac {e^x - 1} x - 1}\) | \(=\) | \(\ds \size {\frac {e^x - 1 - x} x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {\frac {\sum_{n \mathop = 0}^\infty \frac {x^n} {n!} - 1 - x} x }\) | Definition of Exponential Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {x \sum_{n \mathop = 2}^\infty \frac {x^{n-2} } {n !} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size x \sum_{n \mathop = 2}^\infty \frac 1 {n !}\) | as $\size x \le 1$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \size x \sum_{n \mathop = 0}^\infty \frac 1 {n !}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size x e\) | Definition of Euler's Number as Limit of Series | |||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | as $x \to 0$ |
$\blacksquare$