How to prove, that $e^x$ is uniformly continuous if $x$ is negative?

Question

How can one show with only elementary mathematics, that $e^x$ is a uniformly continuous function on $(-\infty;0]$ I started with $\mid e^x-e^y \mid$, knowing that I assume , that $\mid x-y \mid<\delta$, but I see no connection between $\mid e^x-e^y\mid and \mid x-y\mid $ Should I assume, that $y<x$ and try to estimate something? Thank you

Answer 1

If $x \leq 0$, then $e^{x}$ is uniformly continuous.

Note that for every $\epsilon > 0$ there exists $\delta >0$ such that

$$|e^{x}-e^{y}| = |e^{x}||e^{(y-x)}-1| \leq |e^{(y-x)}-1| < \epsilon,$$

when $|x-y| < \delta$ -- where $\delta$ does not depend on $x$ or $y$ because $e^{x}$ is continuous at $x=0$

Answer 2

You can easily show that f(x) is lipschitz-continuous on $(-\infty,0]$, i.e. there exists $L>0$ such that $|f(x)-f(y)|\leq L|x-y|$. Then u can choose $\delta:=\frac{\epsilon}{L}$ , where L is the "lipschitz-constant".

Answer 3

A sufficient condition for $f: A\subset\mathbb R \rightarrow \mathbb R$ differentiable to be uniformly-continuous is that $|f'(x)|<M< \infty$ in $A$. In this case, since $e^x$ is strictly-increasing, we have that $(e^x)'=e^x<1$ in $(-\infty, 0)$