What justifies using L'Hopital's Rule for $\lim\limits_{x\to1}\frac{x^3-x}{\ln(x-1)}$? 
What justifies using L'Hopitals Rule for $\lim\limits_{x\to1}\frac{x^3-x}{\ln(x-1)}$?

I figured that the limit can be computed like that:$$\lim\limits_{x\to1 }\frac{x^3-x}{\ln(x-1)}=\lim\limits_{x\to1 }\frac{3x^2-1}{\frac{1}{x-1}}=\lim\limits_{x\to 1}( 3x^2-1)(x-1)=0$$ But why can we use L'Hopital? We neither have $0/0$ or $\infty/\infty$
 A: You have misused  L'Hospital's rule but you came up with the correct answer. It is your problem, not L'Hospital's rule problem. 
The $x-1$ which jumped to the top and caused the answer to become $0$ is also to blame for the coincidence. 
A: First of all, do you see that the problem is ill-posed as you can only take a right-hand limit here?. (You cannot approach $0$ from the left as the natural logarithm function is not defined for negative values.)
Now, generally speaking, in order to use L'R, you must initially get the indeterminant form  $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
In your case, when we take the RH-limit of the original function, the numerator approaches $0$ as the denominator approaches $-\infty$. Hence, the answer to a modified version of your problem is:
$$ \lim_{x \to 0^{+}}\frac{x^{3}- x}{\ln (x - 1)} = \frac{0}{\to -\infty} = 0.$$ 
A: Like being stated in others’ comments, when $x\to 1^+$, the top goes to $0$ and the bottom goes to $\infty$, so the limit is just $0$.
However, if you really want to apply L’Hôpital’s Rule, in fact, the conditions ARE satisfied in your question. For the indeterminate form $$\lim_{x\to\bullet}{\frac{f(x)}{g(x)}},$$ where $\lim_{x\to\bullet}{|f(x)|} = \infty$, $\lim_{x\to\bullet}{|g(x)|} = \infty$, the first condition about $f(x)$ can be missed. In the case when only the bottom is going to infinity, L’Hôpital’s Rule is still valid. In other words, your work is correct.
You may see the “General Form” section of
https://en.m.wikipedia.org/wiki/L%27Hôpital%27s_rule
