Upper Bound of Natural Logarithm/Proof 2

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

Fix $x \in \R_{>0}$.

We first show that $\forall n \in \N : n\left({ \sqrt[n]{x} - 1 }\right) < x - 1 $

Case 1: $0 < x < 1$
Suppose $0 < x < 1$.

Then:

Case 2: $x = 1$
Suppose $x = 1$.

Then:

Case 3: $x > 1$
Suppose $x > 1$.

Then:

Thus:
 * $\forall n \in \N : n\left({ \sqrt[n]x - 1 }\right) \leq x - 1$

by Proof by Cases.

Thus:
 * $ \displaystyle \lim_{n \mathop \to \infty} \left({ \sqrt[n]x - 1 }\right) \leq x - 1$

from Limit of Bounded Convergent Sequence is Bounded.

Hence the result, from the definition of $\ln$.