Module 1 Practice Problems: Limits and Continuity
Instructions: Work through each problem on paper first, then reveal the solution to check your work. These problems cover all four lessons in Module 1.
Problem 1
Evaluate limx→5 (x² − 25)/(x − 5).
Solution
Direct substitution gives 0/0. Factor: (x² − 25) = (x − 5)(x + 5).
Cancel: (x + 5). Substitute: 5 + 5 = 10.
Problem 2
Evaluate limx→0 (sin 4x)/(7x).
Solution
Rewrite: (sin 4x)/(7x) = (4/7) · (sin 4x)/(4x). As x → 0, (sin 4x)/(4x) → 1.
Answer: 4/7.
Problem 3
Evaluate limx→0 (√(1 + x) − 1)/x.
Solution
Rationalize: multiply by (√(1+x) + 1)/(√(1+x) + 1). Numerator: (1+x) − 1 = x.
Result: x/[x(√(1+x) + 1)] = 1/(√(1+x) + 1). Substitute x = 0: 1/(1 + 1) = 1/2.
Problem 4
Evaluate limx→∞ (6x² − x + 3)/(2x² + 5x).
Solution
Degrees equal (both 2). Ratio of leading coefficients: 6/2 = 3.
Problem 5
Find all vertical and horizontal asymptotes of f(x) = (2x + 1)/(x − 3).
Solution
Vertical: Denominator zero at x = 3; numerator 2(3)+1 = 7 ≠ 0. VA: x = 3.
Horizontal: Degrees equal, ratio = 2/1 = 2. HA: y = 2.
Problem 6
Let f(x) = { 3x − 1 if x < 2, and x² + c if x ≥ 2 }. Find c so that f is continuous at x = 2.
Solution
Left limit: 3(2) − 1 = 5. Right limit and f(2): 4 + c. Set equal: 4 + c = 5, so c = 1.
Problem 7
Use the IVT to show that x³ + x − 5 = 0 has a root in [1, 2].
Solution
f(x) = x³ + x − 5 is continuous (polynomial). f(1) = 1 + 1 − 5 = −3 < 0. f(2) = 8 + 2 − 5 = 5 > 0.
Since f changes sign, the IVT guarantees a root c in (1, 2).
Problem 8
Evaluate limx→−2 (x³ + 8)/(x + 2).
Solution
Factor sum of cubes: x³ + 8 = (x + 2)(x² − 2x + 4). Cancel (x + 2).
Substitute x = −2: (−2)² − 2(−2) + 4 = 4 + 4 + 4 = 12.
Problem 9
Evaluate limx→0 x² cos(1/x).
Solution
Use the Squeeze Theorem. Since −1 ≤ cos(1/x) ≤ 1, we have −x² ≤ x² cos(1/x) ≤ x².
Both outer limits equal 0 as x → 0, so by the Squeeze Theorem the limit is 0.
Problem 10
Classify each discontinuity of f(x) = (x² − x − 2)/((x − 2)(x + 3)).
Solution
Factor numerator: x² − x − 2 = (x − 2)(x + 1).
f(x) = (x − 2)(x + 1)/[(x − 2)(x + 3)] = (x + 1)/(x + 3) for x ≠ 2.
At x = 2: removable discontinuity (the factor cancels; there is a hole at x = 2, y = 3/5).
At x = −3: the simplified form has denominator zero and numerator −2 ≠ 0. This is an infinite (vertical asymptote) discontinuity.