How to prove continuity of $e^x$.

Question

I simply want a proof that $e^x$ is continuous.

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

1. $e$ 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.

2. 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.

Answer 1

Fix $a>1$ (the case $a=1$ is trivial, and $0<a<1$ can be handled by switching to $a^{-1}>1$). 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.

1. $a^{b/c}$ 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.

2. The function $\frac bc\mapsto a^{b/c}$ is strictly increasing: if $\frac bc<\frac pq$ then $a^{b/c}<a^{p/q}$. 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 $a^b<a^p$; again this is clear.

3. 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 $\gamma=y/x\in\Bbb R_{>1}$, and let $n\in\Bbb N$ be sufficiently large that $\gamma^n>a$. Then $y^n>ax^n$ so there is some integer power $a^m$ of $a$ with $x^n<a^m<y^n$. Taking $n$-th roots (again a strictly increasing operation) it follows that one has $x<a^{m/n}<y$, in other words $a^{m/n}\in I$.

4. Any increasing function$~f$ defined on a dense set and without jump discontinuities is continuous. Given $x$ in the domain of $f$, and $\epsilon>0$, find values $x_<,x_>$ with $f(x_<)\in(f(x)-\epsilon,f(x))$ and $f(x_>)\in(f(x),f(x)+\epsilon)$ by point$~$3, and take $\delta=\min(x-x_<,x_>-x)$. Then for all $x'$ with $|x-x'|<\delta$ one has $x'\in(x_<,x_>)$ so $f(x')\in(f(x_<),f(x_>))$ by the increasing property of$~f$, and $(f(x_<),f(x_>))\subset(f(x)-\epsilon,f(x)+\epsilon)$, so that $|f(x)-f(x')|<\epsilon$.

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 $\sup\{\,f(x)\mid x\in D_{<x_0}\,\}=\inf\{\,f(x)\mid x\in D_{>x_0}\,\}$ (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 $x\mapsto x+[x^3>2]$ (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.

Answer 2

$\mathbf{Result \; 1}$: If $a > 0 $, then $\lim_{n \to \infty } a^{\frac{1}{n}} = 1 $.

If you want a solution to this problem, let me know.

$\mathbf{Result \; 2} $: If $a> 0$, then $f(x) = a^x$ is continuous in $\mathbb{R}$

$\mathbf{Solution}:$ Let $a>0$, and let $\epsilon >0$ .We will employ the sequential characterization of continuity. This means that for any given sequence $(x_n) \subseteq \mathbb{R}$ with $x_n \to x$, then $f(x_n) \to f(x)$. To show this, let $(x_n) \subseteq \mathbb{R}$ be an arbitrary sequence such that $x_n \to x$. Notice by arithmetic,

$$ a^{x_n} - a^x = a^x(a^{x_n-x} - 1) $$

By previous result, we can obtain an $N \in \mathbb{N}$ such that for all $n > N$,

$$ |a^{\frac{1}{n}} - 1 | < \frac{\epsilon}{a^x} $$

Similarly, since $x_n \to x$, we can select $K \in \mathbb{N}$ such that for all $n > K$,

$$ |x_n - x| < \frac{1}{N} $$

Therefore, if we put $M = \max\{N,K\} $, we have

$$ |f(x_n) - f(x) | = |a^{x_n} - a^x| = a^x|a^{x_n-x} - 1| < a^x \frac{ \epsilon}{a^x} = \epsilon$$

Hence, by definition, $f(x_n) \to f(x) $. In particular, $f(x) = a^x$ is continuous as desired.

$\mathbf{Proposition}:$ $\exp(x) $ is continuous on $\mathbb{R}$.

$\mathbf{Solution}:$ Apply $\mathbf{Result \; 2}$ to $a = e$, and we are done.

Answer 3

(Imho) the easiest (but maybe to theory burdened) way, I guess, is to define $e^x =\lim_{n\rightarrow\infty}\sum^n_{k=0}\frac{x^k}{k!}$, to show that the series converges locally uniformly and to use the fact that (locally) uniform convergence of continuous functions produces a continuous function.

Answer 4

$f$ 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.

Answer 5

Consider the function $f:(0,\infty)\to\mathbb R$ such that $f(x)=\int_1^x\frac{\mathrm dt}{t}$. This is continuous, differentiable and, in fact $f'(t)=1/t$ for all $t\in(0,\infty)$. In particular, it is clear that $f'(t)>0$ for all such $t$ so that $f$ is strictly increasing in its domain. Some work will show that $f$ is not bounded neither above not below —indeed, and for example, the first is equivalent to the statement that the integral $\int_1^\infty\frac{\mathrm dt}{t}$ diverges.

Now we have

If $h:(a,b)\to(c,d)$ is a continuous, strictly increasing function which is surjective, then there is an inverse function $h^{-1}:(c,d)\to(a,b)$ and it is itself also continuous and strictly increasing.

This applies to our $f$, and shows that it has an inverse function $f^{-1}:\mathbb R\to(0,\infty)$ which is continuous. Since this is of course the exponential function, we have shown what we wanted.

Later. Notice that one can define the exponential functio in this way —and it is done so classically, too. In any case, It is very easy to check that it is the exponential. Let $h=f^{-1}$. A change of variables shows that $f(xy)=f(x)+f(y)$, so applying inverse functions we see that $h(x+y)=h(x)h(y)$. It follows that $(h(x+y)-h(x))/y=h(x)(h(y)-1)/y$ and, since we know that $h$ is differentiable, it follows that $h'(x)=h'(0)h(x)$ for all $x$. The inverse function theorem allows us to compute $h'(0)=1$, so $h'(x)=h(x)$ and $h(0)=1$. One can now compute the derivative of $h(x)e^{-x}$, see that it is identically zero, adn evaluate his function at $0$ to conclude that $h(x)=e^{-x}$.

Answer 6

Your idea of defining exponential function $a^{x}$ is intuitive but has many complications. The definition assumes that $a^{p}$ is defined for all $a > 0$ and $p \in \mathbb{Q}$. If $x \in \mathbb{R} - \mathbb{Q}$ and if $x_{n}$ is a sequence of rationals with $\lim_{n \to \infty}x_{n} = x$ then you define $a^{x} = \lim_{n \to \infty}a^{x_{n}}$. This requires us to first justify that the definition is unambiguous. This further implies that we must establish

1) if $x_{n}$ is a sequence of rationals with $\lim_{n \to \infty}x_{n} = x$ then the limit $\lim_{n\to\infty}a^{x_{n}}$ exists.

2) if $x_{n}, y_{n}$ are sequences of rationals with $\lim_{n \to \infty}x_{n} = \lim_{n \to \infty}y_{n} = x$ then $\lim_{n \to \infty}a^{x_{n}} = \lim_{n \to \infty}a^{y_{n}}$

Both the above results although seem trivial but are not so easy to prove. Assuming that you have overcome the above stumbling blocks in your definition it is easy to show that $a^{x}$ is continuous for all $x$. This requires us to prove the exponential rule $a^{x + y} = a^{x}a^{y}$ which is easy according to the definition adopted here. And then it needs to be established that $\lim_{x \to 0}a^{x} = 1$.

The limit can be established provided we establish the weaker inequality $a^{x} \leq a^{y}$ for $x < y, a > 1$. For $a < 1$ the inequality sign is reversed. This is also easily done by the definition used here. (Hint: If $a > 1$ and $x < y$ then put $h = y - x > 0$ and show that $a^{h} \geq 1$ and then multiply the inequality by $a^{x}$)

Next we need to first show that $a^{1/n} \to 1$ as $n \to \infty$. Let $a > 1$ and we put $a^{1/n} = 1 + b$ so that $a =(1 + b)^{n} > 1 + nb$ so that $$0 < b < \frac{a - 1}{n}$$ Letting $n \to \infty$ we see that $b \to 0$ and hence $a^{1/n} = 1 + b \to 1$.

We can now show that $a^{x} \to 1$ as $x \to 0$. We will consider only the case for $x \to 0^{+}$. Let $\epsilon > 0$ be given. Then there is a positive integer $m $ such that $1 - \epsilon < a^{1/n} < 1 + \epsilon$ for $n \geq m$. Let $\delta = 1/m > 0$. If $0 < x < \delta$ then $0 < x < 1/m $ and then $1 \leq a^{x} \leq a^{1/m}$ so that $1 - \epsilon < a^{x} < 1 + \epsilon$ for $0 < x < \delta$. It follows that $a^{x} \to 1$ as $x \to 0^{+}$.

Now we can see that $$\begin{aligned}\lim_{x \to b}a^{x} &= \lim_{h \to 0}a^{b + h}\\ &= \lim_{h \to 0}a^{b}a^{h} = a^{b}\cdot 1 = a^{b}\end{aligned}$$ so that $a^{x}$ is continuous for all $x$. The proofs above assume that $a > 1$. The case $a < 1$ can be handled by noting that $1/a > 1$ and applying the results on $1/a$.

Update: If you are only concerned why $2^{1/2}\approx 2^{47/99}$ it is easy to provide a simple proof and avoid all limit machinery mentioned above. You ideally want $a^{x}$and $a^{y}$ to be nearly equal if $x, y$ are nearly equal and let's keep $x, y$ rational to avoid all limit arguments. By the way this is in reality the proof for stumbling block 2) which I mentioned in start of my answer.

We have $a^{x} - a^{y} = a^{y}(a^{x - y} - 1)$ so the difference $a^{x} - a^{y}$ will be small if $a^{x - y} - 1$ is small. Let $x - y = h$ and keep $x > y$ otherwise we interchange roles of $x, y$. Then we see that $h$ is small (as $x$ is near to $y$) and we want to make $a^{h} - 1$ small. We have the inequality $$\frac{a^{h} - 1}{h} < a - 1$$ for rational $a, h$ with $a > 1, 0 < h < 1$. See its proof here. Hence we have $0 < a^{h} - 1 < h(a - 1)$. Now we can see that $h$ is small so $h(a - 1)$ is small and therefore $a^{h} - 1$ is small and therefore $a^{x} - a^{y}$ is small.

Answer 7

Okay, I'll be a little stubborn here and I'll try to convince you that defining $e^x$ as a power series is actually better. (Respecting your second wish, I'll not enter in a lot of technicalities)

First, let me say something: People usually think that mathematicians do things in order to complicate simple things. The fact is that most of times, the mathematician is actually simplifying things. As a rule of thumb, in order to simplify something, you need to make a theory, and the more you want to simplify, the more abstract or hard to grasp the theory will be.

Now, okay, let's make a function called $\xi (x)$, defined by:

$\displaystyle \xi (x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ (Here, we adopt $0^0=1$ to avoid notational complications)

Now, don't be afraid of this definition (I didn't even say that this is the exponential! This is a auxiliary function, for now)

The important things are:

1. $\xi$ is continuous. (This follows from it being a power series)
2. $\xi (x+y) = \xi(x)\xi(y)$. (This follows from the Cauchy Product Formula)
3. $\xi$ is defined for ALL reals. (This is a simple observation, but will be extremely useful)
4. $\xi(0)=1$, $\xi(1)=e$

And now, let's take a look at the function you call "exponential". Let's call it $\exp$. We have:

1. $\exp(x+y)=\exp(x)\exp(y)$ for all $n \in \mathbb{N}$ (I hope this is clear)

2. $\exp(0)=1$, $\exp(1)=e$

Now, for $n \in \mathbb{N}$, we have that $\exp(n)=\exp(1)...\exp(1)$ $n$ times=$\exp(1)^n$

For $-n$ negative integer, we have that $1=\exp(0)=\exp(n-n)=\exp(n)\exp(-n)$, hence $\displaystyle \exp(-n)=\frac{1}{\exp(n)}=\frac{1}{\exp(1)^n}=\exp(1)^{-n}$

For $a=\frac{p}{q} \in \mathbb{Q}$, $\exp(p)=\exp(q.\frac{p}{q})=\exp(\frac{p}{q}+\frac{p}{q}+...+\frac{p}{q})$ $q$ times $=\exp(p/q)^q$. Hence, $\exp(\frac{p}{q})=\exp(p)^{1/q}=(\exp(1)^p)^{1/q}$

Now, note the following:

• I computed the values of $\exp(x)$ for all $x$ rational using only properties that $\xi$ AND $\exp$ have in common. Hence, $\xi$ coincides with $\exp$ in the rationals.
• By the previous observation, we have immediately that $\exp$ is continuous in the rationals, as a restriction of a continuous function is continuous.

Okay, now, since we have that $\xi$ is defined for ALL reals, we could very well say:

Well, make $\xi$ be what we will call $\exp$ for the reals, then.

But we can do more. Since $\exp$ coincides with $\xi$ in a dense set ($\mathbb{Q}$), if we want it to extend continuously, the ONLY way is for the extension to be $\xi$ (maybe this is not clear for you, but it is a theorem, that is stated as follows, and holds in a way more general situation:)

Two continuous functions that are equal in a dense subset of $\mathbb{R}$ are equal at the whole domain.

Now, notice the irony. You began asking why $\exp$ is continuous, and we derived quite "easily" that $\exp$ is continuous in the rationals, and the motivation came as to whether we could have a continuous "$\exp$" in the reals.

Just to close up, you may feel like I'm cheating in requiring a continuous extension. Well, I'm not. Following your intuition, I'd say that the only things that you consider the exponential surely satisfies is $\exp(x+y)=\exp(x).\exp(y)$, $\exp(0)=1$ and $\exp(1)=e$. Well, there are a LOT of non-continuous functions satisfying this in $\mathbb{R}$, coinciding with the values of $\exp$ in the rationals. And, by the way, your way of defining $\exp$ for reals as limits of sequences is blalantly, explictly requiring it to be continuous in $\mathbb{R}$.

EDIT: Appendix: Why considering THIS series:

Whatever $\exp$ will be, we want that $\exp(x+y)=\exp(x).\exp(y)$

Now, let's say we should guess a power series for $\exp(x)$

$$\exp(x)=\sum_{n=0}^{\infty}a_nx^n$$

Well, $\displaystyle \exp(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the binomial theorem.

But, hey, look! The right hand side is begging for the cauchy product formula. If that $a_n.n!$ were $1$... then, the right hand side would be $\displaystyle \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{m=0}^{\infty}\frac{y^m}{m!}\right)$. So, why not? Let $a_n.n!$ be $1$, then! So, we have $\displaystyle a_n=\frac{1}{n!}$. Now, look back at the series for $\exp$, and perceive our luck.

Answer 8

Let $a \gt 1$.

Aspiring math students will be able to accept the following assertions.

(1) $f: \Bbb Q \to \Bbb R$ defined by $f(x) = a^x$ is a strictly increasing function
and
(2) $f(u+v) = f(u)f(v)$

These two facts are used to develop the following:

Proposition 1: The decreasing sequence $a^{+{\frac{1}{2^n}}}$ converges to $1$ as $n\to\infty$.
Proof: Exercise, but see just keep hitting that square root key.

Proposition 2: The increasing sequence $a^{-{\frac{1}{2^m}}}$ converges to $1$ as $m\to\infty$.
Proof: Exercise.

Proposition 3: The function $f$ is continuous at 0.
Proof: Exercise.

Proposition 4: The function $f$ is continuous at all points in $\Bbb Q$.
Proof: Exercise.

Proposition 5: For any $q \in \Bbb Q$, $f(q) = sup(f(p))$ with all $p \lt q$.
Proof: Exercise.

Now extend the function $f$ to any real number $x$ by taking the supremum $sup(f(p))$ with $p \in \Bbb Q$ and $p \lt x$. We use the same notation, $f$, for the new function.

Proposition 6: $f: \Bbb R \to (0, +\infty)$ is strictly increasing and surjective.
Proof: Exercise.

Proposition 7: $f(x) = a^x$ is continuous.
Proof: See Prop 6 implies Prop 7.

Proposition 8: The extension $f$ also satisfies (2).
Proof: Exercise (use properties of limits).

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The OP mentioned that they had no intuition for why
$2^{1/2}\approx 2^{47/99}$
But, we can rewrite it with a common denominator,
$2^{99/198}\approx 2^{94/198}$
or
$2^{94/198}2^{5/198}\approx 2^{94/198}$
Now the term on the left is larger, so divide both sides by it,
$1 \approx \frac{2^{94/198}}{2^{94/198}2^{5/198}} = 2^{-5/198} \approx .983 $

So in general, if $0 \lt x \lt y$ and $x$ and $y$ are close you can get the relative (percent) difference between $a^x$ and $a^y$ with

$\frac{a^x}{a^y} = a^{x-y}$