Exp x equals e^x

Theorem

$\exp x = e^x$

where:

$\exp$ is the exponential function

$e$ is Euler's number

$e^x$ is $e$ to the $x$th power

Proof

Let the restriction of the exponential function to the rationals be defined as:

$\exp \restriction_{\Q}: x \mapsto \displaystyle \lim_{n \to +\infty}\left (1 + \frac x n\right)^n$

Let $e$ be Euler's Number defined as:

$e = \displaystyle \lim_{n \to +\infty}\left (1 + \frac 1 n\right)^n$

For $x=0$:

	\(\displaystyle \)	\(\displaystyle \exp \restriction_{\Q} \ 0\)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty}\left (1 + \frac 0 n\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\Box$

For $x \ne 0$:

	\(\displaystyle \)	\(\displaystyle \exp \restriction_{\Q} \ x\)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty}\left (1 + \frac x n\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{\left({n/x}\right) \to +\infty}\left (\left (1 + \frac 1 { \left({n/x}\right) } \right)^{\left({n/x}\right)}\right)^x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent Combination Laws
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• Roland E. Larson,  Robert P. Hostetler and Bruce H. Edwards: Calculus: 8th Edition (2005): $\S 5.5$