Fix $x \in \R_{>0}$. Let $x > 1$. From Power Function on Strictly Positive Base is Convex, $x^h$ is convex. Thus for $0 < h < s$: {{begin-eqn}} {{eqn | l = \frac {x^h - a^0} {h - 0}       | o = \le       | m = \frac {x^s - a^0} {s - 0}       | c = {{Defof|Convex Real Function}} }} {{eqn | ll= \leadsto       | l = \frac {x^h - 1} h       | o = \le       | m = \frac {x^s - 1} s }} {{end-eqn}} Further, $0 < \dfrac 1 x < 1$. So, for $h < s < 0 \iff 0 < -s < -h$: {{begin-eqn}} {{eqn | l = \frac {\paren {\frac 1 x}^{-s} - \paren {\frac 1 x}^0} {-s - 0}       | o = \le       | m = \frac {\paren {\frac 1 x}^{-h} - \paren {\frac 1 x}^0} {-h - 0}       | c = Power Function on Strictly Positive Base is Convex }} {{eqn | ll= \leadsto       | l = \frac {x^s - 1} {-s}       | o = \le       | m = \frac {x^h - 1} {-h}       | c = Exponent Combination Laws: Negative Power }} {{eqn | ll= \leadsto       | l = \frac {x^h - 1} h       | o = \le       | m = \frac {x^s - 1} s       | c = Order of Real Numbers is Dual of Order of their Negatives }} {{end-eqn}} Hence $\dfrac {x^h - 1} h$ is increasing on $\R \setminus \set 0$. Next: {{begin-eqn}} {{eqn | l = h       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = x^h       | o = >       | r = 1       | c = Power Function on Base Greater than One is Strictly Increasing }} {{eqn | ll= \leadsto       | l = x^h - 1       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = \frac {x^h - 1} h       | o = >       | r = 0       | c = Order of Real Numbers is Dual of Order of their Negatives }} {{end-eqn}} So $\dfrac {x^h - 1} h$ is strictly positive on $\R_{>0}$. In particular: :$\dfrac {x^h - 1} h$ is bounded below (by $0$) and increasing on $\openint 0 \to$ :$\dfrac {x^h - 1} h$ is bounded above (by $\displaystyle \inf_{h \mathop > 0} \frac {x^h - 1} h$) and increasing on $\openint \gets 0$ So from Limit of Increasing Function, $\displaystyle \lim_{h \mathop \to 0^+} \frac {x^h - 1} h$ and $\displaystyle \lim_{h \mathop \to 0^-} \frac {x^h - 1} h$ exist. Further: {{begin-eqn}} {{eqn | l = \lim_{h \mathop \to 0^+} \frac {x^h - 1} h       | r = \lim_{h \mathop \to \infty} h \paren {x^{1 / h} - 1}       | c = Limit of Composite Function }} {{eqn | r = \lim_{n \mathop \to \infty} n \paren {x^{1 / n} - 1}       | c = Limit of Sequence is Limit of Real Function }} {{eqn | r = \ln x       | c = Sequential definition of natural logarithm }} {{end-eqn}} where $\sequence {n \paren {x^{1 / n} - 1 } }_{n \mathop \in \N}$ is now a real sequence. Similarly: {{begin-eqn}} {{eqn | l = \lim_{h \mathop \to 0^-} \frac {x^h - 1} h       | r = \lim_{h \mathop \to 0^+} -\frac {x^{-h} - 1} h       | c = Limit of Composite Function }} {{eqn | r = -\lim_{h \mathop \to \infty} h \paren {x^{-1 / h} - 1}       | c = Limit of Composite Function }} {{eqn | r = -\lim_{h \mathop \to \infty} h \paren {\paren {\frac 1 x}^{1 / h} - 1}       | c = Exponent Combination Laws: Negative Power }} {{eqn | r = -\lim_{n \mathop \to \infty} n \paren {\paren {\frac 1 x}^{1 / n} - 1}       | c = Limit of Sequence is Limit of Real Function }} {{eqn | r = -\ln \frac 1 x       | c = {{Defof|Natural Logarithm|subdef = Positive Real|index = 3}} }} {{eqn | r = \ln x       | c = Logarithm of Reciprocal }} {{end-eqn}} Thus, for $x > 1$: {{begin-eqn}} {{eqn | l = \lim_{h \mathop \to 0^-} \frac {x^h - 1} h       | r = \ln x       | c =  }} {{eqn | r = \lim_{h \mathop \to 0^+} \frac {x^h - 1} h       | c =  }} {{end-eqn}} So from Limit iff Limits from Left and Right, for $x > 1$: :$\displaystyle \lim_{h \mathop \to 0} \frac {x^h - 1} h = \ln x$ Suppose instead that $0 < x < 1$. From Ordering of Reciprocals: :$\dfrac 1 x > 1$ Thus, from above: {{begin-eqn}} {{eqn | l = \ln \frac 1 x       | r = \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h }} {{eqn | ll= \leadsto       | l = \ln x       | r = -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h       | c = Logarithm of Reciprocal }} {{eqn | r = -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} {-h}       | c = Limit of Composite Function }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} h }} {{eqn | r = \lim_{h \mathop \to 0} \frac {x^h - 1} h       | c = Exponent Combination Laws: Negative Power }} {{end-eqn}} Hence the result. {{qed}} Category:Natural Logarithms mwvx3z0ccwp1mp2c0tgi4kvsnp6qn4t