I have the following limit: $$\lim_{\frac{1}{x}\rightarrow 0}\frac{\ln\frac{1}{x}-\ln1}{(\frac{1}{x}+1)-1}$$
Imagine we declared $u:=\frac{1}{x}$. Would this limit be the definition of $f'(u)$, or $[f(u)]'=f'(u)\cdot u'$ (because of the chain rule) as u tends towards $0$ and $x$ towards $+\infty$? For better comparison, here are $f'(u)$ and $[f(u)]'$:
$$f'(u)|{_{x=+\infty}}=\ln'(\frac{1}{x}+1)|{_{x=+\infty}}=\frac{1}{\frac{1}{x}+1}|{_{x=+\infty}}=1$$
$$[f(u)]'|{_{x=+\infty}}=f'(u)\cdot u'|{_{x=+\infty}}=\ln'(\frac{1}{x}+1) \cdot (\frac{1}{x})|{_{x=+\infty}}=\frac{1}{\frac{1}{x}+1}\cdot (-\frac{1}{x^2})|{_{x=+\infty}}=0$$
Any help would be appreciated, thank you for your time.
It depends on what derivative you mean with the prime. To make it clear, write $$ \frac{d}{du}f(u)=\lim_{h\to0}\frac{f(u+h)-f(u)}{h} $$ and, if $u$ is a dependent variable of the independent $x$, $$ \frac{d}{dx}f(u)=\frac{d}{du}f(u)·u'(x) $$
The usual convention is that in $f'(g(x))$ the derivative is of $f$ alone at the point $u=g(x)$. $(f(g(x)))'$ does not have a strong conventional definition, but I think most would understand it as total derivative wrt. the independent $x$.
However, $[f(u)]'$ has to be considered as ambiguous. It is not clear from the symbols in that expression that you intend $u$ to be a function of $x$, as it is it would parse as a function in $u$ with derivative for $u$. Or perhaps as the derivative of a constant.
The difference quotient however is for, in the limit, $f'(1)$ with $f(u)=\ln u$.