Let $\delta>0$ be given and $f: A \subset \mathbb{R} \rightarrow \mathbb{R}$ an uniformly continuous function (with $A$ a nonempty interval (?)).
Is it possible to find a $c >0$ such that for all $x,y \in A$ we have
$|x-y| < \delta \Rightarrow |f(x) - f(y) | < c$?
This holds for example for each linear function (which are uniformly continuous). But it does not hold for the exponential function, which is not uniformly continuous.
To elaborate Did's comment:
Let $A\subseteq \Bbb R$ be an interval $f\colon A\to\Bbb R$ be uniformly continuous. By uniform continuity, there exists $a>0$ such that for all $x,y\in A$ with $|x-y|<a$, we have $|f(x)-f(y)|<1$.
For given $\delta>0$, let $n=\left\lceil \frac{\delta}a\right\rceil$ (so that $n-1<\frac \delta a\le n$). Then we succeed with $c=n$:
Let $x,y\in A$ with $|x-y|<\delta$. For $k=0,\ldots, n$, let $t_k=x+\frac kn(y-x)$. Then $t_0=x$, $t_n=y$ and $|t_{k+1}-t_k|=\frac{|y-x|}{n}<\frac \delta n\le a$ for $0\le k<n$. Note that $t_k\in A$ for all $k$ (this is where we use that $A$ is an interval!) We conclude $|f(t_{k+1})-f(t_k)|<1$ and so $$|f(y)-f(x)|\le |f(t_n)-f(t_{n-1})|+\ldots+|f(t_1)-f(t_0)|<n.$$