Function with Limit at Infinity of Exponential Order Zero

Theorem
Let $f:\left[ { 0 \,. \, . \, \to }\right) \to \R$ 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 $\left [{0 \,.\,.\, \to} \right)$.

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

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

Proof
Denote $\displaystyle L = \lim_{t \mathop \to +\infty} f \left({t}\right)$.

Define the constant mapping:


 * $C \left({t}\right) = - L$

Further define:


 * $g \left({t}\right) = f \left({t}\right) + C \left({t}\right)$

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.

From Limit at Infinity, there exists $c \in \R$ such that:
 * $\forall t > c: \left\vert{f \left({t}\right) - L}\right\vert < \epsilon$

Therefore:

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