# 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

Consider the natural logarithm $\ln x$, which is defined on the open interval $\openint 0 {+\infty}$.

$\ln x$ is strictly increasing.
the inverse of $\ln x$ always exists.

The inverse of the natural logarithm function is called the exponential function, which is denoted as $\exp$.

Thus for $x \in \R$, we have:

$y = \exp x \iff x = \ln y$

implies the following definition:

### As the Limit of a Sequence

The exponential function can be defined as the following limit of a sequence:

$\exp x := \displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$

## 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 $\sequence {x_n}$ does indeed converge to a limit.

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

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

It will next be shown that:

$\displaystyle \map \ln {\lim_{n \mathop \to \infty} \sequence {x_n} } = x$

We have:

 $\displaystyle \map \ln {\paren {1 + \frac x n}^n}$ $=$ $\displaystyle n \, \map \ln {1 + x n^{-1} }$ Logarithms of Powers $\displaystyle$ $=$ $\displaystyle x \frac {\map \ln {1 + x n^{-1} } } {x n^{-1} }$ multiplying by $1 = \dfrac {x n^{-1} } {x n^{-1} }$

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:

$\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$

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

Thus:

$\displaystyle x \frac {\map \ln {1 + x n^{-1} } } {x n^{-1} } \to x$

as $n \to \infty$.

$\paren {1 + \dfrac x n}^n = \exp \paren {n \, \map \ln {1 + \dfrac x n} } \to \exp x = e^x$

as $n \to \infty$.

$\blacksquare$