Upper Bound of Natural Logarithm/Proof 2

Proof
Let $\sequence {f_n}$ denote the sequence of mappings $f_n: \R_{>0} \to \R$ defined as:
 * $\map {f_n} x = n \paren {\sqrt [n] x - 1}$

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

We first show that $\forall n \in \N : n \paren {\sqrt [n] x - 1} < 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:
 * $\forall n \in \N: n \paren {\sqrt [n] x - 1} \le x - 1$

by Proof by Cases.

Thus:
 * $\displaystyle \lim_{n \mathop \to \infty} \paren {\sqrt [n] x - 1 } \le x - 1$

from Limit of Bounded Convergent Sequence is Bounded.

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