University of Utah — MATH 1210 Calculus I — Final Exam 150+ Practice Questions with Answers & Rationales

University of Utah — MATH 1210 Calculus I — Final
Exam
150+ Questions with Answers & Rationales


COURSE TOPICS COVERED

Topic Chapters

Limits & Continuity 1.1–1.7

Derivatives 2.1–2.9

Applications of Derivatives 3.1–3.9

Integrals 4.1–4.6

Applications of Integrals 5.1–5.6



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


1. Evaluate lim𝑥→2 (3𝑥 2 − 2𝑥 + 1).

, • A) 7
• B) 9
• C) 11
• D) 13
• E) 15
Answer: B
Rationale: Since the function is a polynomial, it is
continuous everywhere. Direct substitution: 3(2)2 − 2(2) +
1 = 12 − 4 + 1 = 9.


sin(3𝑥)
2. Evaluate lim𝑥→0 .
𝑥

• A) 0
• B) 1
• C) 2
• D) 3
• E) Does not exist
Answer: D
sin 𝑢
Rationale: Using the standard limit lim𝑢→0 =
𝑢
sin(3𝑥) sin(3𝑥)
1: lim𝑥→0 = 3 ⋅ lim𝑥→0 = 3(1) = 3.
𝑥 3𝑥

, 4𝑥 2 +3𝑥
3. Evaluate lim𝑥→∞ .
2𝑥 2 −5

• A) 0
• B) 1
• C) 2
• D) 4
• E) ∞
Answer: C
Rationale: Divide numerator and denominator
4+3/𝑥 4
by 𝑥 2 : → = 2.
2−5/𝑥 2 2



tan 𝑥
4. Evaluate lim𝑥→0 .
𝑥

• A) 0
• B) 1
• C) ∞
• D) Does not exist
• E) −1

, Answer: B
sin 𝑥 tan 𝑥 sin 𝑥 1
Rationale: tan 𝑥 = , so = ⋅ → 1 ⋅ 1 = 1.
cos 𝑥 𝑥 𝑥 cos 𝑥



1−cos 𝑥
5. Evaluate lim𝑥→0 .
𝑥2

• A) 0
1
• B)
2

• C) 1
• D) 2
• E) ∞
Answer: B
𝑥2
Rationale: Using the identity 1 − cos 𝑥 ≈ for small 𝑥, or
2
1−cos 𝑥 sin 𝑥
apply L'Hôpital's Rule twice: lim𝑥→0 = lim𝑥→0 =
𝑥2 2𝑥
1
.
2



ln 𝑥
6. Evaluate lim𝑥→∞ .
𝑥

• A) 0
• B) 1