# Euler's Number: Limit of Sequence implies Base of Logarithm

## Theorem

Let $e$ be Euler's number defined by:

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

Then $e$ is the unique solution to the equation $\ln \left({x}\right) = 1$.

That is:

- $\ln \left({x}\right) = 1 \iff x = e$

## Proof

First we prove that $e$ is a solution to $\ln \left({x}\right) = 1$:

\(\displaystyle \ln \left({e}\right)\) | \(=\) | \(\displaystyle \ln \left({\lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n}\right)\) | $\quad$ Definition:Euler's Number/Limit of Sequence | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \left({\ln \left({1 + \frac 1 n}\right)^n}\right)\) | $\quad$ Sequential Continuity is Equivalent to Continuity in Metric Space | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \left({n \ln \left({1 + \frac 1 n}\right)}\right)\) | $\quad$ Logarithm of Power | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \frac {\ln \left({1 + \frac 1 n}\right)} {1 / n}\) | $\quad$ Inverse of Group Inverse | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{x \mathop \to \infty} \frac {\ln \left({1 + \frac 1 n}\right)} {1 / x}\) | $\quad$ Limit of Sequence is Limit of Real Function | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{x \mathop \to \infty} \frac{\left({-\frac 1 {x^2} }\right) \left({1 + \frac 1 x}\right)} {-\frac 1 {x^2} }\) | $\quad$ L'Hôpital's Rule | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{x \mathop \to \infty} \frac x {x + 1}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ |

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 $\ln \left({x}\right) = 1$ is unique.

Hence the result.

$\blacksquare$