Function of exponential order

Remarks

Example: $f(t) = t^n$ with $n \in \N$ is of exponential order $1$, since $e^t \ge t^n / n!$ for $t \ge 0$ by the power series of $e^t$; so $\abs{f(t)} \le n!\, e^t$ with $T = 0$. In fact $t^n$ is of exponential order $a$ for every $a > 0$. Non-example: $f(t) = e^{t^2}$ is not of exponential order, since $e^{t^2}/e^{at} = e^{t(t - a)} \to \infty$ for every real $a$. A bounded function, such as $\sin t$, is of exponential order $0$. Some books require the bound for all $t \ge 0$ ($T = 0$); for functions that are bounded on bounded intervals the two versions agree, after enlarging $M$.