Limit $\lim_{x \to 1} \frac{\log{x}}{x-1}$ without L'Hôpital

In THIS ANSWER and THIS ONE I showed, without the use of calculus, that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1$$

Therefore, we can write for $x>1$

$$\frac1x \le \frac{\log(x)}{x-1}\le 1$$

and for $x<1$

$$1 \le \frac{\log(x)}{x-1}\le \frac1x$$

whereupon applying the squeeze theorem yields the result $1$.

To elaborate on copper.hat's comment, it does not require L'Hopital's rule to recognize that

$$\lim_{x\to1}{\log x\over x-1}=\lim_{x\to1}{\log x-\log1\over x-1}=f'(1)\quad\text{where }f(x)=\log x$$

This is simply the definition of the derivative. If you also know that $f'(x)=1/x$, then you get

$$\lim_{x\to1}{\log x\over x-1}=1$$

It might be noted that L'Hopital's rule is often invoked for problems like this, but its invocation is purely a matter of convenience.

Rewrite this using $t = h-1$ as $$\lim_{t\to 0} \frac{\log(1+t) - 0}{t}.$$

Now try to recognize this as the derivative of a function you know.

The point of this exercise is almost certainly to recognize that this is the definition of the derivative.