Differentiability of |ln(x)| at x = 1

For $x$ slightly greater than $1$, $f(x) = \ln(x)$, so the right-hand limit is $\lim_{x\to 1^+} \frac{\ln(x) - 0}{x - 1}$. This is exactly the limit definition of the derivative of $\ln(x)$ at $x=1$, which equals $1/1 = 1$. For $x$ slightly less than $1$, $f(x) = -\ln(x)$, so the left-hand limit is $\lim_{x\to 1^-} \frac{-\ln(x) - 0}{x-1} = -\lim_{x\to 1^-}\frac{\ln(x)}{x-1} = -1$.

Since the right-hand limit ($1$) and the left-hand limit ($-1$) of the difference quotient disagree, the two-sided limit $\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}$ does not exist. Therefore $f$ is not differentiable at $x=1$. Graphically, this happens because the two linear pieces $\ln(x)$ and $-\ln(x)$ meet at $x=1$ with different slopes, producing a sharp corner (like the graph of $|x|$ at $0$) rather than a smooth tangent line.