University of California — MATH 1A Calculus — Common Final Exam 150+ Practice Questions with Answers & Rationales

University of California — MATH 1A
Calculus — Common Final Exam
150+ Practice Questions with Answers
& Rationales

PART I: LIMITS & CONTINUITY (Questions 1-30)


1. Evaluate lim𝑥→1 (4𝑥 3 − 𝑥 2 + 2).
• A) 3
• B) 4
• C) 5
• D) 6
• E) 7
Answer: C
Rationale: Since the function is a polynomial, it is
continuous everywhere, so we can evaluate by

,direct substitution: 4(1)3 − (1)2 + 2 = 4 − 1 +
2 = 5.


cos(2𝑥 2 )−1
2. Evaluate lim𝑥→0 .
4𝑥

• A) 0
• B) −1/4
• C) 0
• D) 1
• E) Does not exist
Answer: A
Rationale: Using L'Hôpital's Rule or the known
cos 𝑢−1
limit lim𝑢→0 = 0: as 𝑥 → 0, 2𝑥 2 → 0, and
𝑢
the numerator goes to 0 faster than the
denominator, so the limit is 0.


4 𝑥
3. Evaluate lim𝑥→∞ (1+ ) .
𝑥

, • A) 1
• B) 𝑒
• C) 𝑒 4
• D) 𝑒 1/4
• E) ∞
Answer: C
Rationale: Recall lim𝑧→∞ (1 + 1/𝑧)𝑧 = 𝑒.
Rewrite: lim𝑥→∞ (1 + 4/𝑥)𝑥 = lim𝑥→∞ [(1 +
𝑥/4 4
4/𝑥) ] = 𝑒4.


sin(3𝑥)
4. Evaluate lim𝑥→0 .
𝑥

• A) 0
• B) 1/3
• C) 1
• D) 3
• E) Does not exist

, Answer: D
Rationale: Using the standard
sin(3𝑥)
limit lim𝑢→0 sin(𝑢)/𝑢 = 1: lim𝑥→0 =
𝑥
sin(3𝑥)
lim𝑥→0 3 ⋅ = 3 ⋅ 1 = 3.
3𝑥



3𝑥 2 +2𝑥
5. Evaluate lim𝑥→∞ .
5𝑥 2 −1

• A) 0
• B) 3/5
• C) 5/3
• D) 1
• E) ∞
Answer: B
Rationale: Divide numerator and denominator
3+2/𝑥 3
by 𝑥 2 : → as 𝑥 → ∞.
5−1/𝑥 2 5