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Examen
MA 102 Mock Midterm Exam | Questions and Solutions | 2026 Update | 100% Correct - Wilfrid Laurier University
NuevoPart 1: Limits 1. limx→2x2−4x−2limx→2x−2x2−4 equals: • A) 0 • B) 2 • C) 4 • D) Does not exist Answer: C Rationale: Factor numerator: (x−2)(x+2)(x−2)(x+2). Cancel x−2x−2, leaving limx→2(x+2)=4limx→2(x+2)=4. 2. limx→0sin(3x)xlimx→0xsin(3x) equals: • A) 0 • B) 1 • C) 3 • D) Does not exist Answer: C Rationale: Using the standard limit limθ→0sinθθ=1limθ→0θsinθ=1, rewrite as 3⋅sin(3x)3x→3⋅...
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Examen
MA 102 Mock Final Exam | Questions and Answers | 2026 Update | 100% Correct - Wilfrid Laurier University.
PopularPart 1: Limits & Continuity 1. limx→∞3−1/x2limx→∞23−1/x equals: • A) 0 • B) 3223 • C) ∞∞ • D) Does not exist Answer: B Rationale: As x→∞x→∞, 1/x→01/x→0, so (3−0)/2=3/2(3−0)/2=3/2. 2. limx→0tanxxlimx→0xtanx equals: • A) 0 • B) ∞∞ • C) 1 • D) Does not exist Answer: C Rationale: This is a standard limit: limx→0sinxx=1limx→0xsinx =1 and tanx=sinx/cosxtanx=sinx/cosx, so the limit is 1. 3...
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