Exponential Function is Continuous/Real Numbers/Proof 1

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Theorem

The real exponential function is continuous.

That is:

$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$

Proof

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

Let:

$\ds \exp x = \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n$

Fix $x_0 \in \R$.

Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.

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

From Exponential Sequence is Uniformly Convergent on Compact Sets:

$\paren {1 + \dfrac x n}^n$ is uniformly convergent on $I$.

By the Uniform Limit Theorem:

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n = \exp x$

is continuous on $I$.

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

$\blacksquare$