Equivalence of Definitions of Real Exponential Function/Inverse of Natural Logarithm implies Limit of Sequence

Theorem
The following definition of the concept of the real exponential function:

As the Inverse of the Natural Logarithm
implies the following definition:

Proof
Let $\exp x$ be the real function defined as the inverse of the natural logarithm:
 * $y = \exp x \iff x = \ln y$

Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as:
 * $x_n = \paren {1 + \dfrac x n}^n$

First it needs to be noted that $\left \langle {x_n} \right \rangle$ does indeed converge to a limit.

From Equivalence of Definitions of Real Exponential Function: Limit of Sequence implies Sum of Series, we have:


 * $\displaystyle \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$

From Series of Power over Factorial Converges, the is indeed shown to converge to a limit.

It will next be shown that:
 * $\displaystyle \ln \paren {\lim_{n \mathop \to \infty} \left \langle {x_n} \right \rangle} = x$

We have:

From Limit of Sequence is Limit of Real Function, we can consider the differentiable analogue of the sequence.

From Derivative of Logarithm at One we have:
 * $\displaystyle \lim_{x \mathop \to 0} \frac {\ln \paren {1 + x} } x = 1$

But $x n^{-1} \to 0$ as $n \to \infty$ from Sequence of Powers of Reciprocals is Null Sequence.

Thus:
 * $\displaystyle x \frac {\ln \paren {1 + x n^{-1} } } {x n^{-1} } \to x$

as $n \to \infty$.

From Exponential Function is Continuous:
 * $\paren {1 + \dfrac x n}^n = \exp \paren {n \ln \paren {1 + \dfrac x n} } \to \exp x = e^x$

as $n \to \infty$.