TITLE: Is this Epsilon-Delta approach to prove that $e^x$ is continuous correct?
QUESTION [5 upvotes]: I couldn't find an epsilon-delta proof for continuity of $e^x$ so here's my take:
Suppose $|x - x_0| < \delta$ and fix $\epsilon >0$
Consider $|e^x - e^{x_0}| < \epsilon$, then
\begin{gather}
-\epsilon < e^x - e^{x_0} < \epsilon \\
e^{x_0} - \epsilon < e^x < e^{x_0} + \epsilon \\
ln(e^{x_0} - \epsilon) < x < ln(e^{x_0} + \epsilon) \\
ln(e^{x_0} - \epsilon) - x_0 < x - x_0 < ln(e^{x_0} + \epsilon) - x_0
\end{gather}
Hence $e^x$ is continuous at $x_0$ as long as
$$
\delta = min\{x_0 - ln(e^{x_0} - \epsilon), ln(e^{x_0} + \epsilon) - x_0\}
$$
Since $x_0$ is arbitrarily chosen then $e^x$ is continuous and we are done.
Is this correct, or is it not sufficient enough since it does not cater for uniform continuity?

REPLY [7 votes]: It looks okay, although there are a few things one can do to it. You also have to be careful with taking logs if $e^{x_0}-\epsilon<0$...
I would suggest the following: you have
$$ \lvert e^x-e^{x_0} \rvert = e^{x_0} \lvert e^{x-x_0}-1 \rvert < \varepsilon $$
Now, you have the elementary inequality
$$ e^y \geqslant 1+y, $$
which is easy to prove with even the definition as $\lim_{n \to \infty} (1+y/n)^n$. Substituting $y \to -y$, you have
$$ e^{-y} \geqslant 1-y, $$
and then for $y<1$, both sides are positive, so dividing gives
$$ \frac{1}{1-y} \geqslant e^y, \quad (y<1) $$
Hence
$$ \lvert e^y-1 \rvert \leqslant \max{\left\{|y|,\left\lvert \frac{y}{1-y} \right\rvert\right\}}, \quad (y<1) $$
and then all you have to do is take $\delta$ small enough that $y=x-x_0$ satisfies
$$ \max{\left\{|y|,\left\lvert \frac{y}{1-y} \right\rvert\right\}}<e^{-x_0}\varepsilon $$
and $|y|<1$.
(Oh, and this also shows that the exponential is not uniformly continuous, since you can't get rid of the $x_0$-dependence in a uniform way)