University of Michigan MATH115 Calculus I Final Exam With Questions And Rationalized Answers

University of Michigan MATH115 Calculus I
Final Exam With Questions And Rationalized
Answers

Topic 1: Limits and Continuity
sin 𝑥
Q1. Evaluate lim𝑥→0 .
𝑥

• A) 0
• B) 1
• C) ∞
• D) Does not exist
• E) −1
Answer: B) 1
Rationale: This is the standard special limit. As 𝑥 →
0, sin 𝑥 ≈ 𝑥, so the ratio approaches 1.
sin 𝑥
Q2. Evaluate lim𝑥→∞ .
𝑥

• A) 0
• B) 1
• C) ∞

, • D) Does not exist
• E) −1
Answer: A) 0
Rationale: The numerator oscillates between -1 and 1, while
the denominator grows without bound. By the squeeze
theorem, the limit is 0.
Q3. Which of the following limits are equal to 0? (Select all
that apply.)
𝑥 3 −4𝑥+7
• i. lim𝑥→0
𝑥 4 +2𝑥
𝑥2
• ii. lim𝑥→∞
𝑒𝑥

• iii. lim𝑥→0 ∣ 𝑥 ∣
𝑥 4 +2𝑥
• iv. lim𝑥→0
𝑥 3 −4𝑥+7
𝑥2
• v. lim𝑥→∞
𝑒𝑥

Answer: ii, iii, and v
Rationale:
• i: As 𝑥 → 0, numerator → 7, denominator → 0, so limit
is infinite (not 0).
• ii: Exponential growth dominates polynomial, so limit =
0.

, • iii: ∣ 𝑥 ∣→ 0.
• iv: As 𝑥 → 0, numerator → 0, denominator → 7, so limit
= 0.
• v: Same as ii—exponential dominates.
𝑥2
Q4. Find lim𝑥→0 .
1−cos 𝑥

• A) 0
• B) 1
• C) 2
• D) ∞
• E) Does not exist
Answer: C) 2
𝑥2
Rationale: Using the identity 1 − cos 𝑥 ≈ for small 𝑥, the
2
𝑥2
limit is = 2.
𝑥 2 /2

Q5. For what value of 𝑘 is the following function continuous
at 𝑥 = 2?
𝑥 2 + 1, 𝑥<2
𝑓(𝑥) = {
𝑘𝑥 − 3, 𝑥≥2


• A) 2

, • B) 3
• C) 4
• D) 5
• E) 6
Answer: C) 4
Rationale: For continuity at 𝑥 = 2, the left-hand limit must
equal the right-hand limit and the function value. Left: 22 +
1 = 5. Right: 2𝑘 − 3 = 5 ⇒ 𝑘 = 4.
Q6. Which of the following is a vertical
𝑥 2 −4
asymptote of 𝑓(𝑥) = ?
𝑥−2

• A) 𝑥 = −2
• B) 𝑥 = 0
• C) 𝑥 = 2
• D) 𝑥 = 4
• E) No vertical asymptote
Answer: E) No vertical asymptote
(𝑥−2)(𝑥+2)
Rationale: 𝑓(𝑥) = = 𝑥 + 2 for 𝑥 ≠ 2. The
𝑥−2
function has a hole at 𝑥 = 2, not a vertical asymptote.
3𝑥 2 +2
Q7. Find the horizontal asymptote of 𝑓(𝑥) = .
5𝑥 2 −1