# Function with Limit at Infinity of Exponential Order Zero

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## 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:

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 $a = 0$.

$\blacksquare$