Derivative of Exponential at Zero/Proof 4

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 $x \to 0$

$\blacksquare$