Exponential Function is Continuous/Real Numbers/Proof 1

Theorem
The real exponential function is continuous on $\R$.

That is:


 * $\forall x_0 \in \R: \displaystyle \lim_{x \to x_0} \ \exp x = \exp x_0$

Proof
This proof depends on the limit definition of the exponential function.

Let:
 * $\displaystyle \exp(x) = \lim_{n \to \infty} \left({ 1 + \frac{x}{n} }\right)^n$

Fix $x_0 \in \R$.

Consider $I := \left [{x_0 - 1 \,.\,.\, x_0 + 1} \right]$.

From Closed Bounded Subset of Real Numbers is Compact, $I$ is compact.

From Exponential Sequence is Uniformly Convergent on Compact Sets, $\left({ 1 + \dfrac{x}{n} }\right)^n$ is uniformly convergent on $I$.

By the Uniform Limit Theorem, :$\displaystyle \lim_{n \to \infty} \left({ 1 + \frac{x}{n} }\right)^n = \exp$ is continuous on $I$

In particular, $\exp(x)$ is continuous at $x_0$.