Euler's Number: Limit of Sequence implies Base of Logarithm
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Theorem
Let $e$ be Euler's number defined by:
- $\ds e := \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n$
Then $e$ is the unique solution to the equation $\map \ln x = 1$.
That is:
- $\map \ln x = 1 \iff x = e$
Proof
First we prove that $e$ is a solution to $\map \ln x = 1$:
| \(\ds \map \ln e\) | \(=\) | \(\ds \map \ln {\lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n }\) | Definition of Euler's Number as Limit of Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\map \ln {1 + \frac 1 n}^n }\) | Definition of Limit of Real Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \dfrac {n \map \ln {\dfrac {n + 1} n } } 1\) | Logarithm of Power | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {\map \ln {\dfrac {n + 1} n} } {\paren {\dfrac 1 n} }\) | multiplying numerator and denominator by $\dfrac 1 n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac{\paren {-\dfrac 1 {n^2} } \paren {\dfrac n {n + 1} } } {\paren {-\dfrac 1 {n^2} } }\) | L'Hôpital's Rule | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\frac n {n + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
From Logarithm is Strictly Increasing, $\ln$ is strictly monotone.
By Strictly Monotone Mapping with Totally Ordered Domain is Injective it follows that $\ln$ is an injection.
So the solution to $\map \ln x = 1$ is unique.
Hence the result.
$\blacksquare$