Finding the largest constant $C$ such that $|\ln x−\ln y| \geq C|x−y|$ for all $x, y \in (0, 1]$ Find the greatest value of C such that
$|\ln x−\ln y|≥C|x−y|$
for any $x,y∈(0,1]$. What should my approach be? I can't think of much options.

This may not be the most rigorous approach, but I would start by rearranging the equation as $$|\frac{\ln(x)-\ln(y)}{x-y}|\ge C.$$
Since we want this to work for any $x,y$ in our interval we want to find the minimum value of $|\frac{\ln(x)-\ln(y)}{x-y}|$ on the interval $(0,1]$ and let that equal $C$.  This is going to happen as both $x$ and $y$ draw towards 1.  Then let's set $y=1$ and take the limit as $x$ approaches 1:
$$\lim\limits_{x\rightarrow1}\frac{\ln(x)-0}{x-1}.$$
applying l'Hopital's Rule, we get $$\lim\limits_{x\rightarrow1}\frac{1/x}{1}=1$$
