Absolute extrema on a half-open interval

Compute $f'(x) = \frac{1}{x} - 1$. Setting $f'(x) = 0$ gives $\frac{1}{x} = 1$, so $x = 1$, which lies in $(0,5]$ and is the only critical number (note $f'$ is undefined at $x=0$, but that point is not in the domain interval). Since $f'(x) > 0$ for $0 < x < 1$ and $f'(x) < 0$ for $x > 1$, the First Derivative Test shows $f$ has a local maximum at $x=1$, with $f(1) = \ln(1) - 1 = -1$.

Now check the behavior at the endpoints of consideration: at $x = 5$, $f(5) = \ln(5) - 5 \approx -3.39$, which is less than $f(1) = -1$, so the maximum on the interval is at $x=1$. As $x \to 0^+$, $\ln(x) \to -\infty$ while $-x \to 0$, so $f(x) \to -\infty$. This means $f$ decreases without bound near $x=0$, so no absolute minimum exists on $(0,5]$. Therefore the absolute maximum value is $-1$ at $x=1$, and there is no absolute minimum.