Discussion
The function \( f(x) = \dfrac{x^{2}-9}{x-3} \) can be made continuous at \( x = 3 \) by defining \( f(3) \) as
Answer: C
Cancel the common factor away from the point itself:
\[ \frac{x^{2}-9}{x-3} \;=\; \frac{(x-3)(x+3)}{x-3} \;=\; x+3 \quad (x \neq 3). \]
So \( \lim_{x \to 3} f(x) = 6 \), and assigning \( f(3) = 6 \) removes the discontinuity. The quotient itself is \( 0/0 \) at the point, which is why the value must be supplied
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