Derivative of Exponential at Zero

Theorem

Let $\exp x$ be the exponential of $x$ for real $x$.

Then:

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

Proof 1

For all $x \in \R$, we have the following:

• $\exp 0 - 1 = 0$ from Exponential of Zero and One

• $D_x \left({\exp x - 1}\right) = \exp x$ from Sum Rule for Derivatives

• $D_x x = 1$ from Differentiation of the Identity Function.

Having verified its prerequisites, Corollary 1 to L'Hôpital's rule yields immediately:

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

$\blacksquare$

Proof 2

Note that this proof does not presuppose Derivative of Exponential Function.

We use the definition of the exponential as a limit of a sequence:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{\exp h - 1} h\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{\lim_{n \to \infty} \left({1 + \dfrac h n}\right)^n - 1} h\)	\(\displaystyle \)	\(\displaystyle \)	by definition of the exponential
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{\displaystyle \lim_{n \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\)	\(\displaystyle \)	\(\displaystyle \)	Binomial Theorem
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \lim_{n \to \infty} \frac{\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\)	\(\displaystyle \)	\(\displaystyle \)	as $h$ is constant
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \lim_{n \to \infty} \left({ {n \choose 0} 1 - 1 + {n \choose 1} \left({\frac h n}\right)\frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \left({\frac h n}\right)^k \frac 1 h }\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \lim_{n \to \infty}1 + \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k-1} }{n^k}\)	\(\displaystyle \)	\(\displaystyle \)	Powers of Group Elements
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle 1 + h \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac{h^{k-2} } {n^k}\)	\(\displaystyle \)	\(\displaystyle \)	Powers of Group Elements

The right summand converges to zero as $h \to 0$, and so:

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

$\blacksquare$

Proof 3

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {e^x - 1} x\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {e^x - e^0} x\)	\(\displaystyle \)	\(\displaystyle \)	Exponential of Zero and One
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\to\)	\(\displaystyle \)	\(\displaystyle \left.{\dfrac {\mathrm d} {\mathrm dx} e^x}\right \vert_{x \mathop = 0}\)	\(\displaystyle \)	\(\displaystyle \)	as $x \to 0$, from definition of derivative at a point
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle e^x \vert_{x \mathop = 0}\)	\(\displaystyle \)	\(\displaystyle \)	Derivative of Exponential Function
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	Exponential of Zero and One

$\blacksquare$