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 $x-1\ge0 \Rightarrow x \ge 1$ and $x-2 \ne 0 \Rightarrow x \ne 2$. We see that $x=\boxed{1}$ is the smallest integer value that satisfies both conditions.
Question:
Given $f(x) = \frac{\sqrt{x-1}}{x-2}$, what is the smallest possible integer value for $x$ such that $f(x)$ has a real number value?
Answer:
Let's think step by step.

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Given $f(x) = \frac{\sqrt{x-1}}{x-2}$, what is the smallest possible integer value for $x$ such that $f(x)$ has a real number value?
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