Suppose that $a \in \mathbb{R}$. There's an inequality $$\sup_{|x|\le \delta} \frac{|e^{ax} - 1|}{|x|} \le \frac{e^{|a|\delta}}{\delta},$$ see "Testing statistical hypotheses" by E. Lehmann, J.P. Romano, p.50. It's written without proof there, so maybe it's obvious. I know the proof of this fact, but it's not too short. Is there a more simple proof?
My proof:
$h(a) = \sup_{|x|\le \delta} \frac{|e^{ax} - 1|}{|x|} = \sup_{|x|\le \delta} \frac{|e^{-ax} - 1|}{|x|} $ because instead of $x$ we may consider $-x$. Hence $h(a) = h(-a)$ and $h(a) = h(|a|)$ so w.l.o.g. we will suppose that $a \ge 0$. Further $h(a) = \sup_{|x|\le \delta} \frac{|e^{ax} - 1|}{|x|} = \sup_{x \in [0, \delta]} \frac{|e^{ax} - 1|}{|x|}$, because $e^{x} - e^{0} = e^{\zeta} \cdot (x - 0)$ for $\zeta$ between $0$ and $x$ by Mean value theorem and hence negative $x$ are "bad" (if $x<0$ then $\zeta < 0$ and $e^{\zeta} < 1$). Thus $$h(a) = \sup_{x \in [0, \delta]} \frac{|e^{ax} - 1|}{|x|} = \sup_{x \in [0, \delta]} \frac{e^{ax} - 1}{x}.$$ But $ \frac{e^{ax} - 1}{x} = \frac{\sum_{k \ge 0} \frac{(ax)^k}{k!} - 1}{x} = \sum_{k \ge 1} \frac{a^k x^{k-1}}{k!}$ is the sum of nondecreasing functions (for $x \in [0, \delta])$. So $h(a) = \frac{e^{a \delta} - 1}{\delta} \le \frac{e^{a \delta} }{\delta} $.
Let's consider the case $a > 0$, the other case works similarly.
The function $f(x) = e^{ax}$ is convex, which implies that the slope of the line segment from $(0, f(0))$ to $(x, f(x))$ increases with $x$. In particular $$ \frac{f(x)-f(0)}{x-0} \le \frac{f(\delta)-f(0)}{\delta-0} $$ for $x \le \delta$ and $x \ne 0$. This gives $$ \left| \frac{e^{ax} - 1}{x} \right| = \frac{e^{ax} - 1}{x} \le \frac{e^{a\delta} - 1}{\delta} < \frac{e^{a\delta}}{\delta} $$ for $x \le \delta$ and $x \ne 0$.