Choice of $\delta$ for "brute force" proof of continuity of exponential function $e^x$ 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*}
 A: 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.
A: 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


*

*$1+x\leq \exp(x)$ for all $x$

*$\exp(x)<\frac{1}{1-x}$ for all $x<1$.  
From this,

*one obtains the continuity of $\exp$ at $x=0$, i.e. $\lim_{h\rightarrow0}\exp(h)=1=\exp(0)$,

*one obtains the property $\exp(x+y)=\exp(x)\exp(y)$ for any $x,y$.
Most importantly,

*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.

A: We may suppose $0<\varepsilon<e^a$. Notice that:
$$e^a>e^a-\varepsilon\implies a=\ln(e^a)>\ln(e^a-\varepsilon)$$
and
$$e^a+\varepsilon>e^a\implies \ln(e^a+\varepsilon)>\ln(e^a)=a.$$
In particular,
$$a-\ln(e^a-\varepsilon)>0\quad \textrm{and}\quad \ln(e^a+\varepsilon)-a>0.$$
Define
\begin{align*} 
\delta=\min\{a-\ln(e^a-\varepsilon), \ln(e^a+\varepsilon)-a\}>0.
\end{align*}
Then:
\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 $\ln$.
A: 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.
