How to prove continuity of $e^x$. I simply want a proof that $e^x$ is continuous.

I have never really been able to find something satisfying these points:


$e$ is defined to be the limit $\lim_{n\to\infty}\left(1+{1\over n}\right)^n$. This is not essential, and another reasonable definition may be used if that significantly simplifies things. EDIT: In fact, it does not really matter what the base is taken to be. Any number, or a general $a^x$, would be fine.
For positive integers, we define $a^n:=a\times a\times\cdots\times a$ where there are $n$ multiplications. We then let $a^0=1$ and $a^{-n}=1/a^n$. For roots, $a^{1/n}$ is the (principal) number $x$ such that $x^n=a$. For general rationals, $a^{b/c}=(a^{1/c})^b$ is the $c$'th root brought to the $b$'th power. And for irrational numbers $x$, $a^x=\lim{a^n}$ where $n$ is a sequence of rational numbers which converge to $x$.


In other words, I want a proof of the exponential's continuity derived from the "arithmetic" definition itself. The reason I ask is because as a student I never had an intuition for why a number brought to two close exponents would have to yield similar answers. For example, why would $$2^{1/2}\approx 2^{47/99}$$ I thought? After all, one has a $99$'th root to a $47$'th power, and the other is just a square root.

I now know that I was looking for a proof of continuity... and I'm still looking. Please do not use anything advanced - the kid in me still groans when I see people 'define' $e^x$ as a power series, because I know students everywhere (like I was) are getting confused.

Also, I want the proof to be convincing, as in I can sleep well at night, but it need not be headache-inducingly thorough. For example, I didn't derive some of the laws of exponents. I didn't explain what I mean by an irrational number, or prove that the limit must exist and be unique, etc. These you should take as granted - what I want is a straightforward proof of continuity based on the "actual" (!) definition that could be explained to an advanced calculus student frustrated about this question and ready to push themselves to learn - in other words, to my past self.
Fix $a&gt;1$ (the case $a=1$ is trivial, and $0&lt;a&lt;1$ can be handled by switching to $a^{-1}&gt;1$). I'll assume standard facts about $n\mapsto a^n$ for $n\in\Bbb Z$ are known (it is increasing and satisfies the law of exponents). Now to argue the continuity of $\frac bc\mapsto a^{b/c}=(\sqrt[c]a)^b$ on the set of rational numbers $\frac bc$, which is dense in$~\Bbb R$, one can proceed by the following steps.


$a^{b/c}$ is well defined, namely independent of the representation of the rational number $\frac bc$. This means that whenever $\frac bc=\frac pq$, which by definition means $bq=cp\neq0$, one should have $(\sqrt[c]a)^b=(\sqrt[q]a)^p$.
Taking positive integer powers is an injective operation ($x^n=y^n$ implies $x=y$), so we raise both sides to the power $cq$ (assumed positive), so that we must prove $(\sqrt[c]a)^{bcq}=(\sqrt[q]a)^{pcq}$; now we can cancel each root against part of the exponent, and there remains to prove $a^{bq}=a^{cp}$, which is clear.
The function $\frac bc\mapsto a^{b/c}$ is strictly increasing: if $\frac bc&lt;\frac pq$ then $a^{b/c}&lt;a^{p/q}$. Given the previous point, one may assume the fractions have a common denominator $c=q$. Raising both sides to this power (which is also a strictly increasing operation) means it suffices to show $a^b&lt;a^p$; again this is clear.
This increasing function has no jump discontinuities*: for any nonempty open interval$~I$ of positive real numbers, there is some $\frac bc$ with $a^{b/c}\in I$ (a jump discontinuity would give such interval without images). If $I=(x,y)$ put $\gamma=y/x\in\Bbb R_{&gt;1}$, and let $n\in\Bbb N$ be sufficiently large that $\gamma^n&gt;a$. Then $y^n&gt;ax^n$ so there is some integer power $a^m$ of $a$ with $x^n&lt;a^m&lt;y^n$. Taking $n$-th roots (again a strictly increasing operation) it follows that one has $x&lt;a^{m/n}&lt;y$, in other words $a^{m/n}\in I$.
Any increasing function$~f$ defined on a dense set and without jump discontinuities is continuous. Given $x$ in the domain of $f$, and $\epsilon&gt;0$, find values $x_&lt;,x_&gt;$ with $f(x_&lt;)\in(f(x)-\epsilon,f(x))$ and $f(x_&gt;)\in(f(x),f(x)+\epsilon)$ by point$~$3, and take $\delta=\min(x-x_&lt;,x_&gt;-x)$. Then for all $x'$ with $|x-x'|&lt;\delta$ one has $x'\in(x_&lt;,x_&gt;)$ so $f(x')\in(f(x_&lt;),f(x_&gt;))$ by the increasing property of$~f$, and $(f(x_&lt;),f(x_&gt;))\subset(f(x)-\epsilon,f(x)+\epsilon)$, so that $|f(x)-f(x')|&lt;\epsilon$.


This does not yet entirely answer your question, for which it is necessary to show that any function like in 4. can be uniquely interpolated to a continuous increasing function defined on $\Bbb R$. For $x_0$ not in the domain $D$ of $f$ (here an irrational number) one wants to have $f(x_0)=\lim_{x\to x_0, x\in D}f(x)$. To show that the limit exists, one uses that $\sup\{\,f(x)\mid x\in D_{&lt;x_0}\,\}=\inf\{\,f(x)\mid x\in D_{&gt;x_0}\,\}$ (by the absence of jump discontinuities) which is going to be the value of the limit; now the proof that the limit converges can be done by finding values in $D$ sufficiently close to the limit value and some easy but boring estimations, which I will not do here (as per the headache-avoiding request). Also the proof that the extended function is still is continuous and increasing is straightforward.

*This terminology is is not quite proper, as the function on $\Bbb Q
\to\Bbb R$ sending $x\mapsto x+[x^3&gt;2]$ (the brackets meaning $1$ if the condition holds, $0$ otherwise) is continuous on its domain, so it is somewhat unfair to accuse it of having a jump discontinuity, as the above implies. The proper term for what I call "function without jump discontinuities" would be something like "function the closure of whose image is connected" (assuming an increasing function defined on a dense set in both cases), not very attractive.
$f$ is said to be continuous at $a$ if $\lim_{h\to 0}f(a+h)=f(a)$.

In our case:

$\lim_{h\to 0}f(a+h) =\lim_{h\to 0}e^{a+h} = \lim_{h\to 0}e^ae^h = e^a$

This follows because $e^0=1$.

so $e^x$ is continuous at $a$ for all $a \in \mathbb R$

Doesn't get any simpler than that.