Derivative of Exponential Function/Proof 5

Proof
This proof assumes the limit definition of $\exp$.

So let:
 * $\forall n \in \N: \forall x \in \R: \map {f_n} x = \paren {1 + \dfrac x n}^n$

Let $x_0 \in \R$.

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

Let:
 * $N = \ceiling {\max \set {\size {x_0 - 1}, \size {x_0 + 1} } }$

where $\ceiling {\, \cdot \,}$ denotes the ceiling function.

From Closed Real Interval is Compact, $I$ is compact.

From Chain Rule for Derivatives:
 * $\dfrac \d {\d x} \map {f_n} x = \dfrac n {n + x} \map {f_n} x$

Lemma
From the lemma:
 * $\forall x \in I: \sequence {\dfrac \d {\d x} \map {f_{n + N} } x}$ is increasing

Hence, from Dini's Theorem, $\sequence {\dfrac \d {\d x} f_{n + N} }$ is uniformly convergent on $I$.

Therefore, for $x \in I$:

In particular:
 * $\dfrac \d {\d x} \exp x_0 = \exp x_0$