Evaluating a 0/0 limit with L'Hopital's Rule

Substituting $x=1$ gives $\ln(1) = 0$ in the numerator and $1-1=0$ in the denominator, so the limit has the indeterminate form $0/0$. This confirms L'Hopital's Rule applies, since both $f(x)=\ln(x)$ and $g(x)=x-1$ are differentiable near $x=1$.

Differentiate the numerator and denominator separately: $f'(x) = \frac{1}{x}$ and $g'(x) = 1$. Then $\lim_{x\to 1} \frac{\ln(x)}{x-1} = \lim_{x\to 1} \frac{1/x}{1} = \frac{1}{1} = 1.$ Thus the original limit equals $1$.