Module 1 Quiz: Limits and Continuity
Quiz Instructions
Answer each question, then reveal the answer to check your work. Aim for at least 7 out of 10 correct.
1. Evaluate limx→4 (x² − 16)/(x − 4).
Factor: (x−4)(x+4)/(x−4) = x+4. At x = 4: 8.
2. Evaluate limx→0 (1 − cos x)/x².
Multiply numerator and denominator by (1 + cos x): [1 − cos² x]/[x²(1 + cos x)] = sin²x/[x²(1+cos x)] = [(sin x)/x]² · 1/(1+cos x) → (1)² · 1/2 = 1/2.
3. Find limx→∞ (4x³ + x)/(2x³ − 7).
Degrees equal (3). Ratio of leading coefficients: 4/2 = 2.
4. Find all vertical asymptotes of g(x) = (x + 2)/[(x − 1)(x + 4)].
Denominator zero at x = 1 and x = −4. Numerator nonzero at both. VAs: x = 1 and x = −4.
5. Let f(x) = { x + 3 if x < 1, and 2x + k if x ≥ 1 }. Find k so f is continuous at x = 1.
Left limit: 1 + 3 = 4. Right limit: 2(1) + k = 2 + k. Set equal: 2 + k = 4, k = 2.
6. True or False: If limx→a f(x) exists and f(a) is defined, then f is continuous at a.
False. We also need limx→a f(x) = f(a). If they are not equal, f has a removable discontinuity.
7. Evaluate limx→0 (tan x)/x.
(tan x)/x = (sin x)/(x cos x) = [(sin x)/x] · [1/cos x] → 1 · 1 = 1.
8. Evaluate limx→−∞ (3x + 1)/(x² + 2).
Degree of numerator (1) < degree of denominator (2). The limit is 0.
9. Classify the discontinuity of h(x) = (x² − 9)/(x − 3) at x = 3.
Factor: (x−3)(x+3)/(x−3) = x + 3 for x ≠ 3. The limit is 6 but h(3) is undefined. This is a removable discontinuity (hole at (3, 6)).
10. Use the IVT: does cos x = x have a solution in [0, 1]?
Let f(x) = cos x − x. f(0) = 1 > 0. f(1) = cos 1 − 1 ≈ −0.46 < 0. f is continuous and changes sign, so by the IVT there exists c in (0, 1) with cos c = c. Yes.