How to prove $e^x$ is continuous We know that $$e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...$$ For all $|x|<1$ By algebra of continuous function we can prove $e^x$ is continuous .but but 1. how to prove it if $|x|>1$?
Also 2.how to use $\epsilon$ and $\delta$ method to prove $e^x $ is continuous? Thanks
First, it is easy to show from the series representation that the exponential function satisfies the functional equation $e^{x+y}=e^{x}e^{y}$. Then, we have
$$\left| e^{x+h}-e^{x}\right|=e^x\left| e^{h}-1 \right|$$
Now, for $|h|\le 1$ we will show that for any fixed $\epsilon>0$, there exists a number $\delta>0$, that depends on both $\epsilon$ and $x$, such that whenever $|h|<\delta$, $|e^{x+h}-e^{x}|<\epsilon$.
Proceeding, we see that
$$|e^h-1|\le |h| \sum_{n=1}^{\infty}\frac{1}{n!}\le 2|h|$$
Therefore, whenever $|h|<\delta=\min\left(1,\frac{\epsilon}{2e^x}\right)$, $|e^{x+h}-e^{x}|<\epsilon$. And we are done!
In this setting of finite-dimensional normed vector-spaces, differentiability implies continuity. The derivative of the exponential is itself.
First of all, note that $e^x$ is increasing on $(0, + \infty)$.
Then, for all $x,y \in \Bbb{R}$, let $A= \max \{ |x|, |y|\}$. Then $$|e^x-e^y| = \left| \sum_{n \ge 1} \frac{1}{n!} (x^n-y^n) \right| \le \sum_{n \ge 1} \frac{1}{n!} |x^n-y^n| \le \sum_{n \ge 1} \frac{1}{n!} |x-y||x^{n-1}+ \cdots + y^{n-1}| \le$$ $$\le \sum_{n \ge 1} \frac{1}{n!} |x-y|(|x^{n-1}|+ \cdots + |y^{n-1}|) \le \sum_{n \ge 1} \frac{1}{n!} |x-y| nA^{n-1} = e^A|x-y|$$ so that, for all $x$ and for all $0 < \varepsilon <1$, we can call $\delta = \min \{ \varepsilon / e^{|x|+2} , 1 \}$. If $|x-y| < \delta$, recalling that $$\max \{ |x|, |y|\} \le |x|+\delta \le |x|+1$$, then $$|e^x-e^y| \le e^{\max \{ |x|, |y|\}}|x-y| \le e^{|x|+1}\varepsilon / e^{|x|+2} = \frac{e^{|x|+1}}{e^{|x|+2}} \varepsilon< \varepsilon$$ so that our function is continuous at $x$.
NOTE: here I use only the definition, without using the fundamental property $e^{a+b}=e^ae^b$, neither using $e^x >0$.
With an approach similar to Mark's one, once we prove the property $e^{x+y}=e^{x}\cdot e^{y}$ through the series definition, the continuity of the exponential function turns out to be a consequence of the following fact:
$$ \lim_{x\to 0}\frac{e^{x}-1}{x}=1.\tag{1}$$
If $x>0$, by the series definition it is trivial that $e^{x}\geq x+1$.
So if $x$ is positive and less than one we have:
$$ 1+x \leq e^{x} \leq \frac{1}{1-x}\tag{2} $$
and
$$ \lim_{x\to 0^+}\frac{e^{x}-1}{x}=1 \tag{3} $$
follows by squeezing. Moreover,
$$ \lim_{x\to 0^-}\frac{e^{x}-1}{x} = \lim_{x\to 0^{+}}\frac{1-e^{-x}}{x} = 1 \tag{4} $$
can be proved in the same way, and $(3),(4)$ prove $(1)$.
To establish its radius of convergence, use the Cauchy-Hadamard formula: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem
To prove that $e^{x}$ is actually analytic in every disc, see the proof (too long to reproduce here) of Proposition 2.1 from Cartan's "Elementary Theory of Analytic Functions": https://books.google.com/books?id=xUHDAgAAQBAJ&pg=PA241&dq=cartan+elementary+theory+analytic+functions&hl=en&sa=X&ved=0ahUKEwifscvqwrfOAhVJLSYKHexvD6QQ6AEIHjAA#v=onepage&q=no%20means%20trivial&f=false
To prove this by "the algebra of continuous functions" we need the starting statement that $f(x) = e^x$ is continuous for all $|x<1|$ Then we use two lemmas:
The product of a constant and a continous function is continuous.
For any positive fixed $L$ A function $F:\Bbb{R} \to \Bbb{R}$ which is continuous on every interval of length $L$ is continuous everywhere.
So take $L = 1$ and $g(x)$ = $e^x$ on $[-\frac12, \frac12]$, and consider an interval of length $L$ centered on some $x_0$. Then $g(x)$ is continuous since in its range $|x| < 1$ and in that interval, $e^x = e^{x_0} g(x-x_0)$ which is continuous by the first lemma, since $e^{x_0}$ is a constant in that range. Sonce $x_0$ is arbitrary the second lemma can be applied, showing $e^x$ is everywhere-continuous.
The $\delta(\epsilon)$ method:
Consider for a given point $x$, and a given $\epsilon$ with $|\epsilon|>0$, $$ \delta = \ln(\epsilon e^{-x} + 1)$$
You can check algebraically that the definition of continuity is met with this $\delta$. Note that $\delta$ depends not only on $\epsilon$ but also on $x$, and indeed you can't find a $\delta$ that works for a given $\epsilon$ for all values of $x$. That says the function $e^x$ is continuous but not uniformly continuous.
From what you learned in algebra
$\sum x^n$ converges if |x|<1
So, what about: $\sum \frac {x^n}{n!}?$
For any $x,$ soon enough $n>x$ and $\frac xn < 1.$ In our series, each term equals its predecessor multiplies by $\frac x{n+1}$ and if $n>x, \frac x{n+1} <1$. This means that we can compare our series to a geometric series that we know converges.
$\sum \frac {x^n}{n!} = \sum_\limits{n = 0}^{N-1} \frac {x^n}{n!} + \sum_\limits{n = N}^\infty \frac {x^n}{n!}$ and $\sum_\limits{n = N}^\infty \frac {x^n}{n!} < \frac {x^N}{N!} \sum r^n$ with $|r|<1$
Is that sufficient to prove continuity?
Every finite polynomial is continuous.
$\forall \epsilon > 0, x,y, \exists N>0, \delta >0, |y-x|<\delta\\
(\sum_\limits{n = 0}^{N-1} \frac {y^n}{n!} - \frac{x^n}{n!})<\frac \epsilon 3$
and from above:
$|\sum_\limits{n = N}^\infty \frac {x^n}{n!}| < \frac \epsilon 3$
$\sum_\limits{n = 0}^{N-1} \frac {y^n}{n!} - \frac{x^n}{n!} + \sum_\limits{n = N}^\infty \frac {y^n}{n!} - \sum_\limits{n = N}^\infty \frac {x^n}{n!} < \epsilon$
$|y-x|<\delta \implies |e^y - e^x| <\epsilon$
