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

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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.

$\blacksquare$