Is this Epsilon-Delta approach to prove that $e^x$ is continuous correct? 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?
 A: 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)
A: One nice thing about $e^x$ is that it has a scale invariance property which reduces continuity everywhere to continuity at any particular point. Suppose $e^x$ is continuous at $0$. Then
$$|e^x-e^{x_0}|=|e^{x-x_0}e^{x_0}-e^{x_0}|=e^{x_0}|e^{x-x_0}-1|<\epsilon$$
if we choose $\delta$ such that $|e^y-1|<\epsilon/e^{x_0}$ for all $|y|<\delta$ (and substituting $y=x-x_0$ to get the above).
To show continuity at $0$, there are different approaches depending on your chosen definition. For the infinite series definition, we can use the squeeze theorem to show $1\le e^x\le\frac1{1-x}$ for $x\in (0,1)$:
$$\frac{x^0}{0!}\le\sum_{n=0}^\infty\frac{x^n}{n!}\le\sum_{n=0}^\infty x^n.$$
And for $x<0$ we can substitute $x\mapsto-x$ to get $1\le e^{-x}\le\frac1{1+x}\implies 1+x\le e^x\le1$ (by taking the reciprocal). (In case it wasn't clear, this gives an explicit $\delta$ to use; from the inequalities it follows that $\delta=\min(1,\epsilon,\frac{\epsilon}{1+\epsilon})=\frac{\epsilon}{1+\epsilon}$ is sufficient.)
