# Function with Limit at Infinity of Exponential Order Zero

## Theorem

Let $f: \hointr 0 \to \to \R$ be a real function.

Let $f$ be continuous everywhere on their domains, except possibly for some finite number of discontinuities of the first kind in every finite subinterval of $\hointr 0 \to$.

Let $f$ have a (finite) limit at infinity.

Then $f$ is of exponential order $0$.

## Proof

Denote $\displaystyle L = \lim_{t \mathop \to +\infty} \map f t$.

Define the constant mapping:

$\map C t = - L$

Further define:

$\map g t = \map f t + \map C t$

From:

Constant Function is of Exponential Order Zero,
Sum of Functions of Exponential Order,

it is sufficient to prove that $g$ is of exponential order $0$.

Fix $\epsilon > 0$ arbitrarily small.

By definition of limit at infinity, there exists $c \in \R$ such that:

$\forall t > c: \size {\map f t - L} < \epsilon$

Therefore:

 $\, \ds \forall t \ge c + 1 : \,$ $\ds \size {\map g t}$ $=$ $\ds \size {\map f t + \map C t}$ $\ds$ $=$ $\ds \size {\map f t - L}$ $\ds$ $<$ $\ds \epsilon$ $\ds$ $=$ $\ds \epsilon \cdot e^0$ Exponential of Zero

The result follows from the definition of exponential order, with $M = c + 1$, $K = \epsilon$, and $a = 0$.

$\blacksquare$