Uniform Continuity of a Function The question is as follows:
Fix any $a>0$ and any $m \in \Bbb N$. Prove that $f\colon \Bbb Q \cap [-m,m] \to \Bbb R$ given by $f(x)=a^x$ is uniformly continuous.
Thanks for any help.
 A: Given $\epsilon \gt 0$ we want to find a $\delta$ such that $\vert a^x - a^t \vert \lt \epsilon$ whenever $0 \leq \vert x - t \vert \lt \delta$ for all $x,t\in[-m,m]$.
Consider the following, 
$$\vert a^x - a^t \vert  = \vert a^t \vert \vert a^{x-t}-1 \vert.$$
Because $a^x$ is continous and $[-m,m]$ is compact we know it has a maximum value which we will denote with $M$. Note that $M \gt 0$. We will break the problem into 2 cases  (I) When $a$ is larger than $1$ and (II) when $a$ is less than $1$.
Case (I):
The following hold,


*

*$a^t \leq M \quad \forall t \in [-m,m]$

*$a^{x-t} \leq a^{\vert x-t \vert} \leq a^\delta $

*There exists a $\delta'\gt 0$ such that $\vert a^\delta-1 \vert \lt \epsilon/M$, this follows from the continuity of the exponential function at $0$.


Therefore we have
$$\vert a^x - a^t \vert  = \vert a^t \vert \vert a^{x-t}-1 \vert \leq M \vert a^{\delta}-1 \vert \leq M \epsilon/M = \epsilon$$
Where we chose $\delta \lt \delta'$.
I leave Case (II) to the reader.
