How to solve $\lim_{x\rightarrow1} \frac{\ln (x)}{x-1}$ using pre-derivative calculus? How can I solve:  $$\lim_{x\rightarrow1} \frac{\ln x}{x-1}$$using pre-derivative Calculus (no logarithm series or L'Hôpital's rule)?
 A: Hint: set $x=e^y$; if $x\to1$, then $y\to0$. The limit is a fundamental one.
A: One way (among many) to define $e$ is as the solution of $\lim_{h\to 0} \frac{e^h -1}{h} = 1$.  I don't know how often people do this, but it's true (and it leads to $\frac{d}{dx}e^x = e^x$).  Then define $\ln$ as the inverse function to $\exp$, where $\exp(x) \equiv e^x$.  Make the substitution $x=e^y$ in your limit, simplify, and use this definition of $e$.
I don't know offhand how to justify this definition of $e$ without using derivatives.  If you can show there exists $t>0$ satisfying $\lim_{h\to 0} \frac{t^h -1}{h} = 1$, then the uniqueness question is easy to solve (and $t=e$).  And if you can show $\lim_{h\to 0} \frac{2^h -1}{h} > 0$ ("2" is just an arbitrary number greater than $1$), then the existence problem is easy to solve.  That leaves the problem of proving $\lim_{h\to 0} \frac{2^h -1}{h} > 0$ without using derivatives.  I'm pretty sure it can be done.  I'm also sure someone in MSE knows how.
EDIT: ALTERNATE SOLUTION:  Define $\ln$ by $\ln x = \int_1^x 1/t\,dt$.  This is actually done fairly often.  Defining the definite (Riemann) integral does not require derivatives.  In that sense this solution will be "pre-derivative".  In reality, almost no one or maybe even no one learns about definite integrals before they learn about derivatives, so calling this solution "pre-derivative" is questionable. Anyway, by sketching $y=1/x$ and using the fact that $1/1=1$, you can see that $\ln x \approx x-1$ for $x$ close to $1$.  You can show your limit actually equals $1$ by using the continuity of the function $1/x$ and an epsilon-delta argument. 
The definition I give of $e$ in the first solution above is missing from Wikipedia's article on $e$.  I don't know if there is a good mathematical reason for omitting it (a good reason would be if it were impossible to prove $\lim_{h\to 0} \frac{2^h -1}{h} > 0$ without using derivatives), if it's not popular enough, or if it's just a careless omission.  Wikipedia isn't perfect.
Both solutions to your problem avoids derivatives, but the whole exercise strikes me as artificial.
A: Let me show you one more answer. Defining $u = x-1$, your limit becomes:
$$L = \lim_{u\to 0} \frac{\ln{(u+1)}}{u} = 1,$$
since $\ln(u+1)$ behaves like $u$ as $u \to 0$, indeed $\ln(u+1) \sim u$ as $u\to0$.
Cheers! 
