\begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\ &= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\ &= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\ \end{align}
\begin{align} &= \lim_{h\to0}{\ln(1 + \frac{h}{x})}^\frac{1}{h} \\ \end{align} Now this is the part I'm asking about, because I see most people set the reciprocal of my limit, but this one seems to work: \begin{align} n=\frac{x}{h}, h=\frac{x}{n}, \frac{1}{h} = \frac{n}{1}*\frac{1}{x} \end{align} So when h tends towards 0, n tends towards infinity. \begin{align} &= \lim_{n\to\infty}{\ln\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\ \end{align} Here we have \begin{align} {\ln\lim_{n\to\infty}\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\ \end{align} \begin{align} =\frac1x\ln(e) = \frac1x \end{align} This is my first post, I was just curious. Please don't berate me if it's a silly question.