Derivative of Exponential at Zero

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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 1

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:

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

$\blacksquare$


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:

\(\ds \frac {\exp h - 1} h\) \(=\) \(\ds \frac {\lim_{n \mathop \to \infty} \paren {1 + \dfrac h n}^n - 1} h\) Definition of Exponential Function
\(\ds \) \(=\) \(\ds \frac {\ds \lim_{n \mathop \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1} h\) Binomial Theorem
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \frac {\ds \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1} h\) as $h$ is constant
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren { {n \choose 0} 1 - 1 + {n \choose 1} \paren {\frac h n} \frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \paren {\frac h n}^k \frac 1 h}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty}1 + \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 1} }{n^k}\) Powers of Group Elements
\(\ds \) \(=\) \(\ds 1 + h \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 2} } {n^k}\) Powers of Group Elements

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

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

$\blacksquare$


Proof 3

\(\ds \frac {e^x - 1} x\) \(=\) \(\ds \frac {e^x - e^0} x\) Exponential of Zero
\(\ds \) \(\to\) \(\ds \valueat {\dfrac \d {\d x} e^x} {x \mathop = 0} {}\) Definition of Derivative of Real Function at Point, as $x \to 0$
\(\ds \) \(=\) \(\ds \valueat {e^x} {x \mathop = 0} {}\) Derivative of Exponential Function
\(\ds \) \(=\) \(\ds 1\) Exponential of Zero

$\blacksquare$


Proof 4

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 $x \to 0$

$\blacksquare$