# Exponential Tends to Zero and Infinity

## Contents

## Theorem

Let $x \in \R$ be a real number.

Let $\exp x$ be the exponential of $x$.

Then:

- $\exp x \to +\infty$ as $x \to +\infty$
- $\exp x \to 0$ as $x \to -\infty$

Thus the exponential function has domain $\R$ and image $\left({0 \,.\,.\, +\infty}\right)$.

## Proof

### The exponential function approaches positive infinity as x approaches positive infinity

Let $M$ be a strictly positive real number.

Let $N$ be $\ln M$.

$N$ is real because $M > 0$.

From Exponential is Strictly Increasing:

- $\forall x: x > N \implies \exp x > \exp N = M$

Therefore:

- $\forall M \in \R_{>0} : \exists N \in \R : \forall x > N : \exp x > M$

The result follows from the definition of infinite limit at infinity.

$\blacksquare$

### The exponential function approaches 0 as x approaches negative infinity

Let $\epsilon$ be a strictly positive real number.

Let $c$ be $\ln \epsilon$.

Let $x$ be any real number that smaller than $c$.

From Exponential of Real Number is Strictly Positive:

- $\exp x > 0 \implies \left\vert{\exp x}\right\vert = \exp x$

From Exponential is Strictly Increasing:

- $\left\vert{\exp x - 0}\right\vert = \exp x < \exp c = \epsilon$

Therefore:

- $\forall \epsilon \in \R_{>0} : \exists c \in \R : \forall x < c : \left\vert{\exp x - 0}\right\vert < \epsilon$

From the definition of limit involving infinity, the result follows.

$\blacksquare$

### The exponential function has domain $\R$ and image $\left({0 \,.\,.\, +\infty}\right)$

We have that the Exponential is Strictly Increasing.

From above, $\displaystyle \lim_{x \to \infty} \exp x = \infty$

From above, $\displaystyle \lim_{x \to -\infty} \exp x = 0$

Hence the result.

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 14.4$