Is this limit dependent on x. Derivative natural log.

Question

$$ \begin{align} \frac{d}{dx}\ln x &= \lim_{h \to 0} \, \frac{\ln{(x+h)} - \ln x}{h} \\ \\ &= \lim_{h \to 0} \, \frac{1}{h} \, {\ln{\bigg(\frac{x+h}{x}\bigg)}} \\ \\ &= \lim_{h \to 0} \, {\ln{\bigg(1+\frac{h}{x}\bigg)}}^{1/h} \\ \\ &= {\ln{\lim_{h \to 0} \, \bigg(1+\frac{h}{x}\bigg)}}^{1/h}, \text{ let } \frac{h}{x} = \frac{1}{n} \tag{*}\\ \\ &= {\ln{\lim_{n \to \infty} \, \bigg [\bigg(1+\frac{1}{n}\bigg)}}^{n} \, \bigg]^{1/x} \tag{A} \\ \\ &= \frac{1}{x} \ln \lim_{n \to \infty} \, \bigg(1+\frac{1}{n}\bigg)^{n} \tag{B} \\ \\ &= \frac{1}{x} \ln e = \frac{1}{x} \end{align}$$

So my question is: is this a legal step from (A) to (B)? I removed $x$ from the limit, but in step (*) I sort of created $ x = h\cdot n$ which is the part that's confusing me. Does this indicate $x$ depends on $n$? Any other tips on the clarity of the proof would be great, thanks!

Answer 1

$x=h\cdot n$ doesn't imply the dependence of $x$ on either of variables. $x$ doesn't depend on $n$. But $n$ does depend on $h$. They both adjust themselves so that they together multiply to $x$.

For example, the famous equation - $$m=\frac{F}{a}$$ (It's rearranged form of $F=ma$)

Someone might conclude that mass of an object increases as force on it increase, but that, obviously, isn't the case. Acceleration increases proportionately to the force, so that their ratio is constant.

Answer 2

In the calculation of $$ \lim_{h \to 0} \, \frac{\ln{(x+h)} - \ln x}{h}, $$ $x>0$ is a fixed real number. The equality $x=h\cdot n$ tells you how $n$ depends on $h$. (I would rather write $x=h\cdot y$ since $n$ usually means an integer.)