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?
Best Answer
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.
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 $x_0$.