jmc

In order for $f(x)$ to have a real number value, the expression inside the square root in the numerator must be non-negative and the denominator must not be 0. So we have the two conditions $x-1\ge 0\Rightarrow x\ge 1$ and $x-2\ne 0\Rightarrow x\ne 2$. We see that $x=\boxed{1}$ is the smallest integer value that satisfies both conditions.