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 $x \to 0$ |
$\blacksquare$