Q.Find the natural domain of $f(x)=\sqrt{x-2}+\dfrac{1}{x-5}$.

Every part of the rule must give a real value: the square root needs a non-negative radicand and the fraction needs a non-zero denominator.

The rule has two parts, and $f(x)$ is real only where both are defined.

Square-root part $\sqrt{x-2}$: a real square root requires a non-negative radicand:

Fraction part $\dfrac{1}{x-5}$: the denominator must not be zero:

Combine. The domain is the set of $x$ satisfying both conditions — all $x\ge 2$ except $x=5$:

$\text{Domain}=[2,\infty)-\{5\}=[2,5)\cup(5,\infty).$ …

Students often handle one restriction and forget the other — either allowing $x=5$ (division by zero) or writing $x>2$ instead of $x\ge 2$ (the square root of $0$ is pe …