Derivative of Exponential Function/Proof 5

Theorem
Let $\exp$ be the exponential function.

Then:
 * $D_x \left({\exp x}\right) = \exp x$

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

So let:
 * $\displaystyle \forall n \in \N : \forall x \in \R : f_n \left({ x }\right) = \left({ 1 + \frac{x}{n} }\right)^{n}$

Let $x_0 \in \R$.

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

And let $N = \left\lceil{ \max \left\{ { \left\vert{ x_0 - 1 }\right\vert, \left\vert{ x_0 + 1 }\right\vert } \right\} }\right\rceil$, where $\left\lceil{ \cdot }\right\rceil$ denotes the ceiling function.

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

From Chain Rule:
 * $\displaystyle D_x f_n \left({ x }\right) = \frac{ n }{ n + x } f_n \left({ x }\right)$

Lemma
From the lemma:
 * $\displaystyle \forall x \in I : \left\langle{ D_x f_{ n + N } \left({ x }\right) }\right\rangle$ is increasing

Hence, from Dini's Theorem, $\left\langle{ D_x f_{ n + N } }\right\rangle$ is uniformly convergent on $I$.

Therefore, for $x \in I$:

In particular:
 * $D_x \exp x_0 = \exp x_0$