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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?

PinkyWay
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mesllo
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    Your proof, although correct, relies on several facts about $\log$, which (assuming it is defined as the inverse of $\exp$) is usually shown to be a total function using the intermediate value theorem and hence assuming that $\exp$ is continuous! – Mario Carneiro Apr 06 '15 at 13:36
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    So in a sense, to be fully complete I have to properly define $e^x$ in the context of $log(x)$? – mesllo Apr 06 '15 at 13:44
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    As with many $\exp$ and $\log$ proofs, to be formally correct and noncircular you have to be specific about which of the several competing definitions you intend for these functions. The answer I have given below assumes that $e^x:=\sum_{n=0}^\infty\frac{x^n}{n!}$ and $\log x$ is the inverse to $e^x$, and Chappers' answer assumes $e^x=\lim_{n\to\infty}(1+x/n)^n$. The definitions are equivalent, but since the equivalences are somewhat difficult to prove, it changes the approach to "elementary" properties of $e^x$. – Mario Carneiro Apr 06 '15 at 13:49
  • $a<x<b \implies |x|<max{|a|,|b|}$. – Infinite Jan 27 '21 at 12:00

2 Answers2

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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)

Community
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Chappers
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    I am having a hard time putting this proof together. I can see that $y+1 \le e^y \le \frac{1}{1-y}$, which implies $y \le e^y-1 \le \frac{y}{1-y}$ or substituting $y=x-a$, we get $x-a \le e^{x-a}-1 \le \frac{x-a}{1-(x-a)}$. But what would be our choice of $\delta$, and how would it involve $\epsilon$? Thanks. – EthanAlvaree Jul 28 '21 at 19:28
  • I posted a follow-up question, in case you have time to answer. Thanks.

    https://math.stackexchange.com/questions/4211777/choice-of-delta-for-brute-force-proof-of-continuity-of-exponential-function

     – EthanAlvaree Jul 29 '21 at 06:17
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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.)

Mario Carneiro
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