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We aren't allowed to use many tricks such as difference quotient / integral calculus...

Prove that $\exp$ is continuous at $x_{0}=0$

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Given:

$$\exp: \mathbb{R} \ni x \mapsto \sum_{k=0}^{\infty } \frac{1}{k!} x^{k} \in \mathbb{R}$$

also $e = \exp(1)$. For all $x \in \mathbb{R}$ with $\left | x \right | \leq 1$: $$\left | \exp(x) - 1 \right | \leq \left | x \right | \cdot (e-1)$$

and $\exp(0) = 1$

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If I remember correctly, we said that if $|f(x) - f(x_0)| < \varepsilon$ is true then it's continuous.

So I think it would be good to start with: $$e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n$$ then show this is convergent: $$\lim_{x \to x_0} \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = \lim_{x \to x_0} e^x = e^{x_0} = \lim_{n \to \infty} \left(1 + \frac{x_0}{n}\right)^n = \lim_{n \to \infty} \lim_{x \to x_0} \left(1 + \frac{x}{n}\right)^n$$

and in the end put it somehow in $|f(x) - f(x_0)| < \varepsilon$ to show $\exp$ continuous? I don't know exactly how to do that but the way is correct so far?

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    $\begingroup$ It's not clear to me whether you've already proved that for $|x|<1$, $|\exp(x) - 1|\leq |x|(e-1)$. If so you should be able to prove continuity directly from this fact. $\endgroup$
    – Micapps
    Commented Jun 15, 2016 at 13:52
  • $\begingroup$ Yes I did, to be exactly for |x| <=1. But how could I directly proof it using that? It seems like the way I went was too complicated xd $\endgroup$
    – tenepolis
    Commented Jun 15, 2016 at 13:54
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    $\begingroup$ Well, let $\varepsilon>0$. You want to prove that there exists some $\delta>0$ s.t. if $|y-0|<\delta, |\exp(y)-\exp(0)|<\varepsilon$. Since you know that $|\exp(y)-\exp(0)|=|\exp(y)-1|\leq |y|(e-1) <\delta (e-1)$, all you have to ensure is that $\delta (e-1) \leq \varepsilon$. $\endgroup$
    – Micapps
    Commented Jun 15, 2016 at 13:57

2 Answers 2

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Here, we present a proof of the continuity of $e^x$ that relies on elementary tools only, including a basic set of inequalities for the exponential function.

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$.


To show that $e^x$ is continuous at $x_0$ we write

$$\begin{align} e^x-e^{x_0}=e^{x_0}(e^{x-x_0}-1) \tag 2 \end{align}$$

where we used the property $e^xe^y=e^{x+y}$ which I proved in THIS ANSWER.

We restrict $x$ so that $ |x-x_0| < 1$. Then, applying $(1)$ to $(2)$, we find that

$$\begin{align} e^{x_0}(x-x_0)\le e^x-e^{x_0} \le e^{x_0}\frac{x-x_0}{1-(x-x_0)} \end{align}$$

whereby application of the squeeze theorem reveals

$$\lim_{x\to x_0}(e^x-e^{x_0})=0$$

Therefore, $e^x$ is continuous at $x_0$ for all $x_0$.

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  • $\begingroup$ Hi Mark, I really, really like this answer. Is this still provable without Squeeze Theorem? How much extra steps would it take? I opened a separate question and linked to your answer there: math.stackexchange.com/questions/4211777/… $\endgroup$ Commented Aug 2, 2021 at 2:56
  • $\begingroup$ Sure. Restrict $|x-x_0|<1/2$. Then, for any given $\varepsilon>0$, $|e^x-e^{x_0}|<2e^{x_0}|x-x_0|<\varepsilon$ whenever $|x-x_0|<\delta =\frac12 \min(1, e^{-x_0}\varepsilon)$. $\endgroup$
    – Mark Viola
    Commented Aug 2, 2021 at 3:53
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I'm not sure if you are allowed to use this approach, but if it were me, I would take this one. Firstly, I would start by proving that $\frac{1}{x}$ is continuous for $x>0$. Then I will have $ln(x) = \int_1^x{\frac{dt}{t}}$ is continuous, and finally I will have $e^x = ln^{-1}(x)$ is continuous.

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