Question 9 • Precalculus
At x = 2, the denominator x − 2 equals 0, so f(2) is undefined. The numerator also equals 0, and factoring gives (x − 2)(x − 10), so the function simplifies to x − 10 for x ≠ 2; this means the limit exists, but it does not make f(2) exist. Since continuity requires the function to be defined at the point, the function is not continuous at x = 2. Choice B is correct.
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