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Calculus — Medium JEE math MCQ

**[JEE Mains 2026]** If f(3) = 18, f'(3) = 0, and f''(3) = 4, then the value of lim(x→3) [f(x) - 18]/(x - 3)² is:
  1. A. 0
  2. B. 2
  3. C. 4
  4. D. 1

Solution

Using L'Hôpital's rule (0/0 form): lim(x→3) [f(x) - 18]/(x - 3)² = lim(x→3) f'(x)/[2(x-3)] (still 0/0). = lim(x→3) f''(x)/2 = f''(3)/2 = 4/2 = 2.

MATH

medium • PYQ Reworded
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[JEE Mains 2026] If f(3) = 18, f'(3) = 0, and f''(3) = 4, then the value of lim(x→3) [f(x) - 18]/(x - 3)² is:
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Calculus — Medium JEE Mathematics MCQ | MyGoalPrep