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Unit 1 · Limits & Continuity Flashcards Cheat Sheet Essentials Visual Review MC Practice FRQ Practice

AP Calculus AB/BC Unit 1 FRQ Practice

Practice a College Board-style free response question on evaluating limits, continuity, and the Intermediate Value Theorem. Write your response, then reveal the model answer to see exactly what earns each point.

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Free Response Question · Unit 1 · Limits & Continuity

Let f be the function defined by

f(x) = (x² − 9)/(x − 3) for x ≠ 3, and f(3) = 4.

The function g is continuous on the closed interval [0, 5] with g(0) = −1 and g(5) = 7.

A
Evaluate lim(x→3) f(x). Show the algebraic work that supports your answer.

✓ Model answer

Direct substitution gives 0/0, so factor the numerator: (x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3 for x ≠ 3. Therefore lim(x→3) f(x) = 3 + 3 = 6.

Why it scores: Recognizes the indeterminate 0/0 form, factors and cancels correctly, and evaluates the simplified limit to 6. Simply plugging in x = 3 (getting 0/0) without simplifying would not earn credit.
B
Is f continuous at x = 3? Justify your answer using the definition of continuity, and state how f could be redefined to make it continuous there.

✓ Model answer

f is not continuous at x = 3. The definition of continuity requires lim(x→3) f(x) = f(3). Here f(3) = 4 is defined and the limit exists and equals 6, but 6 ≠ 4, so the third condition fails. This is a removable discontinuity; redefining f(3) = 6 (the value of the limit) would make f continuous at x = 3.

Why it scores: States "not continuous," justifies it by comparing the limit (6) to f(3) = 4 against the definition, identifies it as removable, and gives the correct redefinition. A yes/no answer without the definition-based justification would lose credit.
C
Using the given information about g, explain why there must be a value c in the interval [0, 5] such that g(c) = 2.

✓ Model answer

g is continuous on the closed interval [0, 5], and the value 2 lies between g(0) = −1 and g(5) = 7. By the Intermediate Value Theorem, there must exist at least one value c in [0, 5] with g(c) = 2.

Why it scores: Explicitly checks the two IVT hypotheses — continuity on the closed interval AND that 2 is between g(0) and g(5) — before invoking the theorem by name. Omitting either hypothesis, or not naming the IVT, would lose credit.

How to score points on AP Calculus FRQs