Square Root is of Exponential Order Epsilon

Theorem
The positive square root function:


 * $t \mapsto \sqrt t$

is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.

Proof
Recall from Identity is of Exponential Order Epsilon, $t < K'e^{a't}$ for any $a' > 0$, arbitrarily small in magnitude.

Therefore the inequality $\sqrt{t} < K^{at}$ has solutions of the same nature.