I have read several answers (example 1, example 2) about continuity of $e^x$, but most rely on Power Series definition of $e^x$, or sequential definition of a limit, or squeeze theorem.

I would like a brute-force proof that meets the following criteria:

  • Does NOT use sequential definition of limit
  • Does NOT use Squeeze Theorem
  • Uses $\epsilon-\delta$ definition of continuity directly
  • Does NOT use perturbations (e.g. $|e^{a + h} - e^a|$)
  • Uses definition of limit, starting with a $0 < |x-a| < \delta$ and ending with $|e^x - e^a| < \epsilon$
  • Is NOT based on power series definition of $e^x$
  • Is based on elementary limit definition $e^x = \lim_{n \to 0} (1+n)^{\frac{x}{n}}$

I would like to use exponential bounds (which come from Bernoulli's Inequality) like this answer: \begin{align*} y+1 \le \ & \ e^y \le \frac{1}{1-y} \\ \to \quad \quad y \le \ & \ e^y - 1 \ \le \ \frac{y}{1-y} \\ \to \quad x-a \le \ & \ e^{x-a}-1 \ \le \ \frac{x-a}{1-(x-a)} \end{align*} except I am trying to modify that proof so it doesn't depend on Squeeze Theorem.

Proof attempt:

Let $\epsilon > 0$ and $a > 0$ arbitrary. Choose $\delta = \frac{\epsilon}{e^a}$. Then \begin{align*} & \quad 0 < |x - a| < \delta \quad \quad \quad \textrm{ (Given)}\\ &\to \quad |e^{x-a}-1| \quad < \delta \quad \quad \textrm{ (Reason unknown?)} \\ &\to \quad |e^{x-a}-1| < \frac{\epsilon}{e^a} \quad \quad \textrm{ (Substitute $\delta=\frac{\epsilon}{e^a}$)} \\ &\to \quad e^a|e^{x-a}-1| < \epsilon \quad \quad \textrm{ (Multiply both sides by $e^a$)} \\ &\to \quad |e^x-e^a| < \epsilon \quad \quad \quad \textrm{ (Distribute $e^a$ into absolute value)} \\ & \to \quad \lim_{x \to a} e^x = e^a \quad \quad \quad \textrm{ (Definition of limit)} \end{align*}

I know my proof is supposed to use the exponential bounds, $$x-a \le e^{x-a}-1 \le \frac{x-a}{1-(x-a)},$$ so I tried using it (probably incorrectly) in step 2. Just because $|x-a| < \delta$ doesn't mean $e^{x-a}-1$ (bigger) is also less than $\delta$. It may be bigger than $\delta$. So I am having trouble going from step 1 to step 2.

Edit 7/29 ($2^{nd}$ Proof Attempt):

Some comments are suggesting, based on Chappers' answer here, that I should choose $$\delta=\max\left\{|x-a|, \left|\frac{x-a}{1-(x-a)}\right|\right\}.$$ Making this substitution, our proof becomes \begin{align*} & \quad 0 < |x - a| < \delta \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\textrm{ (Given)}\\ & \quad 0 < |x - a| < \max\left\{|x-a|, \left|\frac{x-a}{1-(x-a)}\right| \right\} \quad \quad \quad \textrm{ (Substitution of $\delta$)}\\ &\quad \quad \quad \vdots \\ &\quad \quad \quad ? \\ &\quad \quad \quad \vdots \\ &\to \quad e^a|e^{x-a}-1| < \epsilon \quad \quad \textrm{ (Multiply both sides by $e^a$)} \\ &\to \quad |e^x-e^a| < \epsilon \quad \quad \quad \textrm{ (Distribute $e^a$ into absolute value)} \\ & \to \quad \lim_{x \to a} e^x = e^a \quad \quad \quad \textrm{ (Definition of limit)} \end{align*} I am not sure how to fill in the gaps. The left hand side needs to somehow become $e^a |e^{x-a}-1|$. The right hand side needs to become $\epsilon$. But it seems to me that by making this choice, $\delta$ is no longer a function of $\epsilon$.

Edit 8/19 (Final proof):

For those it helps, here's my final proof based off Paramanand Singh's answer:

Let $\epsilon > 0$ and $a>0$ arbitrary. Choose $\delta= \frac{1}{2}\min\left\{1,\frac{\epsilon}{e^a}\right\}$. Then \begin{align*} & \quad \left|x - a\right| < \delta \tag{Given} \\ \to& \quad \left|x - a\right| < \frac{1}{2} \min\left\{1, \frac{\epsilon}{e^a}\right\} \tag{$\delta = \frac{1}{2}\min\left\{1,\frac{\epsilon}{e^a}\right\}$} \\ \to& \quad 2\left|x - a\right| < \min\left\{1, \frac{\epsilon}{e^a}\right\} \tag{Multiplication by 2} \\ \to& \quad \left|\frac{x-a}{1-(x-a)}\right| < \min\left\{1,\frac{\epsilon}{e^a}\right\} \tag{$\left|\frac{h}{1-h}\right|<2|h|$ if $|h|<\frac{1}{2}$} \\ \to& \quad \left|\frac{x-a}{1-(x-a)}\right| < \frac{\epsilon}{e^a} \tag{$\min\left\{1,\frac{\epsilon}{e^a}\right\}< \frac{\epsilon}{e^a}$} \\ \to& \quad \left|e^{x-a}-1\right| < \frac{\epsilon}{e^a} \tag{Exponential Bound Lemma} \\ \to& \quad e^a\left|e^{x-a}-1\right| < \epsilon \tag{Multiplication by $e^a$} \\ \to& \quad \left|e^a\cdot e^{x-a}-e^a\right| < \epsilon \tag{Distribution Property} \\ \to& \quad \left|e^x-e^a\right| < \epsilon \tag{$e^s\cdot e^t = e^{s+t}$} \\ \to& \quad \lim_{x \to a} e^x = e^a \tag{Definition of limit} \end{align*} The above proof relies on the facts $e^x\cdot e^y=e^{x+y}$ and also the Exponential Bound Lemma $e^x \ge 1+x$, which gives \begin{align*} & \quad e^h \ge \left(1+\frac{h}{n}\right)^n \\ \to& \quad e^h \ge 1+h \\ \to& \quad e^{-h} \ge 1-h \\ \to& \quad e^h \le \frac{1}{1-h} \\ \to& \quad e^h-1 \le \frac{h}{1-h}. \end{align*}

  • $\begingroup$ What you mean under "specifically" in last question? $\delta$ is defined as $\max$ and this gives estimation for exponent i.e. gives desired continuity. $\endgroup$
    – zkutch
    Jul 29, 2021 at 7:35
  • $\begingroup$ Do you mean choose $\delta < \max\left\{x-a, \frac{x-a}{1-(x-a)}\right\} < \frac{\epsilon}{e^a}$ or $\delta = \max\left\{x-a, \frac{x-a}{1-(x-a)}\right\} < \frac{\epsilon}{e^a}$ or $ \max\left\{x-a, \frac{x-a}{1-(x-a)}\right\} < \delta < \frac{\epsilon}{e^a}$, or $ \max\left\{x-a, \frac{x-a}{1-(x-a)}\right\} < \delta = \frac{\epsilon}{e^a}$? $\endgroup$ Jul 29, 2021 at 15:10
  • $\begingroup$ I am having trouble seeing how to use this in the proof. Also, how does $\frac{x-a}{1-(x-a)}$ come into play? Why not choose $\delta = \max\left\{x-a, e^{x-a}-1\right\}$, since I am trying to make sure that $e^{x-a}-1 < \delta$ in step 2 of my proof? $\endgroup$ Jul 29, 2021 at 15:12
  • $\begingroup$ Accordingly linked answer you take $\delta = \max$. How this estimation comes in play is proved again there and why to take it - because it works. $\endgroup$
    – zkutch
    Jul 29, 2021 at 15:58
  • 2
    $\begingroup$ Your definition of $e^x$ is complicated. It would be much better instead to use $e^x=\lim_{n\to\infty} (1+(x/n))^n$. $\endgroup$
    – Paramanand Singh
    Jul 30, 2021 at 7:24

4 Answers 4


This is an expansion of my comments. Using any chosen definition of $e^x$ (eg $e^x=\lim_{n\to\infty} (1+(x/n))^n$) one needs to establish the following two properties of $e^x$:

  • $e^{x+y} =e^xe^y\, \forall x, y\in\mathbb {R} $
  • $e^x\geq 1+x\,\forall x\in(-1,1)$

The second property holds for all real $x$, but it is sufficient (for current question) if one can establish it for the interval $(-1,1)$.

We proceed on the assumption that the above mentioned properties of $e^x$ are proved.

Consider any arbitrary $\epsilon>0$ and let us analyze the target inequality $$|e^{a+h} - e^a|<\epsilon\tag{1} $$ where $a$ is fixed and $h$ is variable. This is equivalent to $$|e^h-1|<\epsilon e^{-a}\tag{2} $$ Next we impose a restriction that $|h|<1/2$. If $0<h<1/2$ then we have $$e^{-h} \geq 1-h$$ or $$e^h\leq \frac{1}{1-h}$$ or $$e^h-1\leq \frac{h} {1-h}<2h\tag{3} $$ If $-1/2<h<0$ then we have $$|e^h-1|=1-e^h=\frac{e^{-h} - 1}{e^{-h}}<e^{-h}-1$$ And using $(3)$ we thus obtain $$|e^h-1|<2(-h)=2|h|$$ Hence we have proved that if $0<|h|<1/2$ then $$|e^h-1|<2|h|\tag {4}$$ Let us now choose $\delta'=\epsilon e^{-a} /2$ and $\delta =\min(1/2,\delta')$. Then for all values of $h$ with $0<|h|<\delta$ we have $$|e^h-1|<2|h|<\epsilon e^{-a} $$ and thus the desired inequality $(2)$ (and equivalently $(1)$) holds. It follows that $e^x$ is a continuous at point $a$.

Some concerns have been raised in comments by Oliver Diaz and presumably these are due to the fact that I haven't presented the details of the definition of $e^x$ suggested in this answer. A simple development of this definition is available in this answer and you may have a look at it.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Aug 30, 2021 at 19:00

This is a rather long comment so I put all this in the answer section.

  • There are a few things to notice form Mark Viola's solution which the OP uses as a template for his argument. The author (Mark) shows that there is a function $\exp:x\mapsto\lim_n\big(1+\frac{x}{n}\big)^n$ defined on the real line that satisfies
  1. $1+x\leq \exp(x)$ for all $x$
  2. $\exp(x)<\frac{1}{1-x}$ for all $x<1$.
    From this,
  3. one obtains the continuity of $\exp$ at $x=0$, i.e. $\lim_{h\rightarrow0}\exp(h)=1=\exp(0)$,
  4. one obtains the property $\exp(x+y)=\exp(x)\exp(y)$ for any $x,y$.
    Most importantly,
  5. the arguments depend only on the inequality $(1+y)^n\geq 1+ny$ for all $y>-1$ and $n\in\mathbb{N}$, which does not involve the the exponential function itself (no circular arguments!)
  • The continuity of $\exp$ at any point $a$ follows easily from this, for $$|e^x-e^a|=e^a|e^{x-a}-1|$$ Since $\exp(h)$ is continuous at $h=0$, $\lim_{x\rightarrow a}e^{x-a}=1$ and so, $\lim_{x\rightarrow a}|e^x-e^a|=0$. (One can write things in terms of $\varepsilon-\delta$ arguments, but it can be avoided since it has been already established that $\exp$ is continuous at $0$.)

There are other (equivalent) ways to introduce the exponential function and obtain its continuity along the way.

  • In modern Calculus texts, the function $\log:x\mapsto\int^x_1\frac{1}{t}\,dt $ is introduce first, continuity, strict monotonicity, differentiability, as well as the known properties of log from antiquity are then established. The exponential function is then defined as the inverse of $\log$ and all desired properties follow by the inverse map theorem.
  • There is a other method I am aware of, that dates back to the German school in the mid-to-late 1800's. There, given a number $a>1$, rational powers $r\mapsto a^r$ are defined, monotonicity established, and using the axiom of supreme (or equivalents) the extension is done for real powers. Continuity at $0$ is established and the rest is as above.
  • $\begingroup$ First, thank you for the answer. I know you say that $\epsilon - \delta$ arguments can be avoided, but I'm really hoping (and the reason I opened this question) was to see a brute-force proof involving a choice of $\delta$ and steps that would show $0 < |x-a| < \delta \quad \to\quad |e^x-e^a| < \epsilon$. If you can flesh out these steps, I would happily accept your answer. Thank you $\endgroup$ Aug 2, 2021 at 0:35
  • $\begingroup$ Thanks Oliver, that makes sense. So you're saying prove continuity at 0 first, e.g. prove $\lim_{h \to 0} e^h = 1 = e^0$ first. But doesn't the argument based on $1+h \le e^h \le \frac{1}{1-h}$ require Squeeze Theorem to be applied? $\endgroup$ Aug 2, 2021 at 2:46
  • $\begingroup$ I see how Squeeze theorem implies continuity of $e^x$. Unfortunately, College Board's curriculum treats Squeeze Theorem after it treats Limit Laws. So the order in which I'm teaching is: Prove continuity of $e^x$ and its inverse $\ln(x)$, then prove limit laws (including continuity of $f(x)^{g(x)}$, whose proof depends on continuity of $\ln(x)$, and after limit laws introduce Squeeze Theorem. Kind of strange order but it's what I have to work with, hence why I have to "brute force" this proof. Is this still provable without Squeeze Theorem, and how much extra steps would it take? Thanks again. $\endgroup$ Aug 2, 2021 at 3:40
  • $\begingroup$ Why would you introduce transcendental functions (exp, log) before the more elementary functions (polynomials, rational functions) I mean, Before the introduction by Wierstrass people has no problem with estimating limits, differentials and even had a notion of continuity, but if one want to proceed with the $\varepsilon-\delta$ formalism, the natural order would be limits of polynomials, rational functions, tricks about limits (including the squeeze theorem) and then trigonometric functions, and the lot of transcendentals. Otherwise, arguments become cir alar and confusing. $\endgroup$
    – Mittens
    Aug 2, 2021 at 4:28
  • $\begingroup$ Hi Oliver, thanks for your comment and sorry for the slow reply. Would the following order (1) limits of polynomial & rational functions, (2) limits of transcendental functions $\sin(x)$, $\cos(x)$, $e^x$, $\ln(x)$, etc., (3) limit laws like $\lim(f+g)$, $\lim(f-g)$, $\lim(fg)$, $\lim(f/g)$, $\lim(f^g)$, and finally (4) tricks like Squeeze Theorem be acceptable? For better or for worse, this is the order College Board directs. $\endgroup$ Aug 6, 2021 at 9:04

We may suppose $0<\varepsilon<e^a$. Notice that:

$$e^a>e^a-\varepsilon\implies a=\ln(e^a)>\ln(e^a-\varepsilon)$$


$$e^a+\varepsilon>e^a\implies \ln(e^a+\varepsilon)>\ln(e^a)=a.$$ In particular,

$$a-\ln(e^a-\varepsilon)>0\quad \textrm{and}\quad \ln(e^a+\varepsilon)-a>0.$$


\begin{align*} \delta=\min\{a-\ln(e^a-\varepsilon), \ln(e^a+\varepsilon)-a\}>0. \end{align*}


\begin{align*} |x-a|<\delta&\implies |x-a|<\min\{a-\ln(e^a-\varepsilon), \ln(e^a+\varepsilon)-a\}\\ &\implies \ln(e^a-\varepsilon)-a<x-a<\ln(e^a+\varepsilon)-a\\ &\implies \ln(e^a-\varepsilon)<x<\ln(e^a+\varepsilon)\\ &\implies e^a-\varepsilon<e^x<e^a+\varepsilon\\ &\implies |e^x-e^a|<\varepsilon. \end{align*}

I adapted this argument from Landau's wonderful book "Differential and Integral Calculus". There are some very nice insights there.

P.s.: I'm looking forward for the analogous question concerning $\ln$.

  • $\begingroup$ Hi PtF, thanks very much for the answer. This is my favorite answer so far, because it captures the spirit of the question, providing a proof that starts with $|x-a|<\delta$ and ending with $|e^x-e^a|<\epsilon$. Before I accept this answer however, I have 2 concerns. This answer looks a lot like the OP from this question: math.stackexchange.com/questions/1222342/… $\endgroup$ Aug 6, 2021 at 9:08
  • $\begingroup$ Concern #1: A commenter to the linked question wrote "you also have to be careful taking logs if $|e^a−\epsilon|<0$" since $\ln(e^a-\epsilon)$ may be undefined, making the proof invalid. Concern #2: Another comment to the linked question that said the proof "relies on several facts about $\ln$, 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!" The commenter recommended instead using Bernoulli's Inequality (which I tried and failed to apply without Squeeze Theorem). $\endgroup$ Aug 6, 2021 at 9:11
  • $\begingroup$ The proof you mention misses some crucial points. In fact, that proof does not ensure $e^a-\varepsilon>0$, whereas mine does, for I start with the assumption $0<\varepsilon<e^a$ what implies $e^a-\varepsilon>0$. As to your second point, I don't see right now why you would need to use continuity of exp to introduce $\ln$. Once $e^x$ is strictly increasing it is injective and therefore a bijection on its image $(0, +\infty)$, the inverse you call $\ln$. $\endgroup$
    – PtF
    Aug 6, 2021 at 12:19

Suppose we have $\varepsilon > 0$. Since $e^x$ is strictly increasing, it is sufficient to show that $\exists\delta > 0$ such that:

$$\mid e^{a+\delta} - e^a\!\!\mid,\; \mid\! e^{a} - e^{a-\delta}\!\mid\; < \varepsilon$$

But note that:

$$\mid e^{a+\delta} - e^a\! \mid \; < \varepsilon $$

$$\iff e^{\delta} \;\cdot \mid\! e^{a} - e^{a-\delta}\!\mid\; < \varepsilon $$

$$\implies \; \mid\! e^{a} - e^{a-\delta}\!\mid\; < \varepsilon e^{-\delta} < \varepsilon $$

since $\delta > 0$. So it actually suffices that there exists $\delta > 0$ such that

$$e^{a+\delta} - e^a < \varepsilon$$

$$\iff e^{\delta} < \bigg(1+\frac{\varepsilon}{e^a}\bigg)$$

Using the inequality from the answer you linked,

$$\impliedby \frac{1}{1-\delta} < \bigg(1+\frac{\varepsilon}{e^a}\bigg)$$

(This is true for $\delta < 1$, which we can assume WLOG.)

And after some algebraic manipulation, this gives that:

$$\delta < \frac{\varepsilon}{e^a + \varepsilon}$$

is sufficient. Let e.g. $\delta = \frac{\varepsilon}{2(e^a + \varepsilon)} > 0$. This is positive for all $a \in \mathbb{R}$, so we deduce that $\exp$ is continuous everywhere.

  • 1
    $\begingroup$ Thank you. Can this method be modified to show $|e^x-e^a| < \epsilon$, instead of $|e^{a+\delta}-e^a|< \epsilon$? Would the same choice of $\delta$ work? $\endgroup$ Jul 29, 2021 at 19:53
  • 1
    $\begingroup$ @EthanAlvaree Yes, this choice of $\delta$ works for all $x$ in the relevant interval -- I explained this in the first half of my answer. Since $\exp$ is strictly increasing, it suffices to consider $x$ at the endpoints of the interval i.e. $x = a \pm \delta$. Then, I further showed that if it works for $x = a + \delta$, it will work for $x = a - \delta$. Thus, to reiterate, all of the values in the interval are covered in proving the case $x = a + \delta$. $\endgroup$ Jul 29, 2021 at 20:12

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