User:Keith.U/Sandbox/Lemma

Theorem
Let $\left\langle{ f_n }\right\rangle$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:
 * $f_n \left({ x }\right) = n \left({ \sqrt[n]{ x } - 1 }\right)$

Let $k \in \N$.

Let $J = \left[{ \dfrac{1}{k}, \,.\,.\, k }\right]$.

Then the sequence of derivatives $\left\langle{ f_n' }\right\rangle$  converges uniformly to some $g : J \to \R$.

Proof
From Derivative of Nth Root and Combination Theorem for Sequences:
 * $\forall n \in \N \forall x \in J : D_x f_n \left({ x }\right) = \dfrac{ \sqrt[n]{x} }{ x }$

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

Thus:

In particular, $\left\langle{ f_n' }\right\rangle$ is convergent to a  continuous function on $J$.

For $x \in \left[{ \dfrac{1}{k}, \,.\,.\, 1 }\right]$:

So $\left\langle{ f_n' \left({ x }\right) }\right\rangle$ is increasing on $\left[{ \dfrac{1}{k}, \,.\,.\, 1 }\right]$.

For $x \in \left[{ 1, \,.\,.\, k }\right]$:

So $\left\langle{ f_n' \left({ x }\right) }\right\rangle$ is decreasing on $\left[{ \dfrac{1}{k}, \,.\,.\, 1 }\right]$.

Thus $\left\langle{ f_n' \left({ x }\right) }\right\rangle$ is monotone on $\left[{ \dfrac{1}{k}, \,.\,.\, 1 }\right] \cup \left[{ 1, \,.\,.\, k }\right] = J$.

From Dini's Theorem, $\left\langle{ f_n' }\right\rangle$ converges uniformly to $\dfrac{1}{x}$ on $J$.

Hence the result.