Proving that the exponential function is continuous

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 $x_0$.