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: {{begin-eqn}} {{eqn | l = 0       | o = <       | m = x       | mo= <       | r = 1 }} {{eqn | ll= \leadsto       | l = 0       | o = <       | m = \sqrt [n] x^{n - k}       | mo= <       | r = 1       | c = Power Function on Base between Zero and One is Strictly Decreasing/Rational Number       | cc= $\forall k \in \set {0, 1, \ldots, n - 1}$ }} {{eqn | ll= \leadsto       | l = 0       | o = <       | m = \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k}       | mo= <       | r = n       | c = Real Number Ordering is Compatible with Addition }} {{eqn | ll= \leadsto       | o =        | m = \frac 1 n       | mo= <       | r = \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }       | c = Ordering of Reciprocals }} {{eqn | ll= \leadsto       | o =        | m = 1       | mo= <       | r = \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }       | c = multiplying both sides by $n$ }} {{eqn | ll= \leadsto       | o =        | m = x - 1       | mo= >       | r = \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }       | c = multiplying both sides by $x - 1 < 0$ }} {{eqn | ll= \leadsto       | o =        | m = y^n - 1       | mo= >       | r = \frac {\paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }       | c = substituting $y = \sqrt [n] x$ }} {{eqn | ll= \leadsto       | o =        | m = y^n - 1       | mo= >       | r = n \paren {y - 1}       | c = Sum of Geometric Sequence }} {{eqn | ll= \leadsto       | o =        | m = x - 1       | mo= >       | r = n \paren {\sqrt [n] x - 1}       | c = substituting $\sqrt [n] x = y$ }} {{end-eqn}} {{qed|lemma}}        === Case 2: $x = 1$ === Suppose $x = 1$. Then: {{begin-eqn}} {{eqn | l = \map \ln x       | r = \map \ln 1 }} {{eqn | r = 0       | c = Natural Logarithm of 1 is 0/Proof 3 }} {{eqn | r = 1 - 1 }} {{eqn | r = x - 1 }} {{end-eqn}} {{qed|lemma}} === Case 3: $x > 1$ === Suppose $x > 1$. Then: {{begin-eqn}} {{eqn | l = x       | o = >       | r = 1 }} {{eqn | ll= \leadsto       | l = \sqrt [n] x^{n - k}       | o = >       | r = 1       | c = Power Function on Base Greater than One is Strictly Increasing/Rational Number       | cc = $\forall k \in \set { 0, 1, \ldots, n - 1}$ }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k}       | o = >       | r = n       | c = Real Number Ordering is Compatible with Addition }} {{eqn | ll= \leadsto       | l = \frac 1 n       | o = >        | r = \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }       | c = Ordering of Reciprocals }} {{eqn | ll= \leadsto       | l = 1       | o = >        | r = \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }       | c = Multiply both sides by $n$ }} {{eqn | ll= \leadsto       | l = x - 1       | o = >       | r = \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }       | c = multiplying both sides by $x - 1 > 1$ }} {{eqn | ll= \leadsto       | l = y^n - 1       | o = >       | r = \frac {n \paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }       | c = substituting $y = \sqrt [n] x$ }} {{eqn | ll= \leadsto       | l = y^n - 1       | o = >       | r = n \paren {y - 1}       | c = Sum of Geometric Sequence }} {{eqn | ll= \leadsto       | l = x - 1       | o = >       | r = n \paren {\sqrt [n] x - 1}       | c = Substitute $\sqrt [n] x = y$ }} {{end-eqn}} {{qed|lemma}} 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 $\ln$. {{qed}}