Continuity proof for exponential Show that $f(x) = e^x$ is continuous using the epsilon-delta definition.
I can't seem to write down anything meaningful...
 A: Let $a$ be a positive real number. Then the function  $f: \mathbb{R}\to \mathbb{R}$ defined by $x\mapsto a^x$ is continuous. 
Proof: 
1) First we prove the continuous at $0$. 
We may assume that $a>1$. Let  $\varepsilon>0$ be arbitrary. Since $a^{1/k}\to 1$ and $a^{-1/k}\to 1$ as $k\to \infty$, we choose $K$  such that both are $\varepsilon$-close to $1$. Let  $\delta=1/K$, so for $|x|<1/K$ we have 
\begin{align}-1/K<x<1/K\\a^{-1/K}<a^{x}<a^{1/K} \end{align}
which proves that $a^{x}$ is $\varepsilon$-close to $1$ as desired (the case $a\le1$ is handled similarly just with the inequality in the other direction).
2) To conclude we prove the continuity in the general case. 
Let  $x_0\in \mathbb{R}$, we have to show that $\lim_{x\to x_0}a^x=a^{x_0}$. Since $x-x_0\to 0$ as $x\to x_0$, then $a^{x-x_0}\to1$ by ($1$) and thus 
$$\lim_{x\to x_0}a^x=\lim_{x\to x_0}a^{x_0}a^{x-x_0}=a^{x_0}\lim_{x\to x_0}a^{x-x_0}=a^{x_0}$$
as was to be shown.
Since $e\in \mathbb{R}^{>0}$, thus $e^x$ is continuous.
A: Start with $|x-c|<\delta$.
Let's take a look at the case that $x\ge c$.
Then:
$$x-c <\delta$$
$$x < c+ \delta$$
$$e^x < e^{c+\delta}$$
$$0 \le e^x-e^c < e^c(e^\delta - 1)$$
Let's pick $\delta$ such that $\varepsilon = e^c(e^\delta - 1)$.
Then:
$$\delta = \ln(1+\varepsilon e^{-c})$$
Repeat for $x<c$.
In other words, you are able to find a $\delta$ for any $\varepsilon>0$.
Qed.
A: Almost identical questions have been answered $n + 1$ times on this site. The proof is trivial. I will present a slight variation of the standard proof.
Recall that, by definition, $f(x)$ is continuous on $\mathbb{R}$ if and only if
$$(\forall x_0 \in \mathbb{R})(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in \mathbb{R})(0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon).$$
Combining a well-worn result
$$e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n,$$
and the fact that, by definition, $\lim_{x \to x_0} f(x) = L$ if and only if
$$(\forall \varepsilon > 0)(\exists \delta > 0)(0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon),$$
proves that $e^x$ is continuous.
Edit: To summarize, we know that the sequence
$$s_n = \left(1 + \frac{x}{n}\right)^n$$
converges to $e^x$ as $n \to \infty$. We see that
$$\lim_{x \to x_0} \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = \lim_{x \to x_0} e^x = e^{x_0} = \lim_{n \to \infty} \left(1 + \frac{x_0}{n}\right)^n = \lim_{n \to \infty} \lim_{x \to x_0} \left(1 + \frac{x}{n}\right)^n.$$
Moreover, by definition, $\lim_{x \to x_0} f(x) = L$ if and only if
$$(\forall \varepsilon > 0)(\exists \delta > 0)(0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon).$$
Yet, since $x_0$ is arbitrary, we have
$$(\forall x_0 \in \mathbb{R})(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in \mathbb{R})(0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon).$$
Thus, $e^x$ is continuous for all $x$.
A: As discussed in this forum, it suffices to show continuity at zero.
Using the fact that $E(0)=1,$ we see that for any $\delta<1,$
every $x\in\mathbb{R}$ (or $x\in\mathbb{C}$) such that $|x|<\delta$ satisfies
\begin{align*}
    \left|E(x) - E(0)\right|
\hspace{1mm}=\hspace{1mm}
    \left|\sum_{n=1}^\infty \frac{x^n}{n!}\right|
\hspace{1mm}\leq\hspace{1mm}
    \sum_{n=1}^\infty \frac{|x|^n}{n!}
\hspace{1mm}<\hspace{1mm}
    \sum_{n=1}^\infty \frac{\delta^n}{n!}
\hspace{1mm}<\hspace{1mm}
    \sum_{n=1}^\infty \delta^n
\hspace{1mm}=\hspace{0.5mm}
    \frac{1}{1-\delta} - 1,
\end{align*}
at which point the reader is invited to choose $\delta$ in terms of a given $\varepsilon$.
