Question 1

Evaluate lim(x→0) [sin(3x)]/x.

Accepted Answer

Setup: This is a standard limit involving the sine function and requires knowledge of the fundamental limit lim(u→0) sin(u)/u = 1.

Solution:
Step 1: Recognize that direct substitution gives 0/0 (indeterminate form).
Step 2: Rewrite the expression: [sin(3x)]/x = 3 · [sin(3x)]/(3x).
Step 3: Let u = 3x. As x → 0, we have u → 0.
Step 4: Apply the fundamental limit: lim(u→0) sin(u)/u = 1.
Step 5: Therefore, lim(x→0) [sin(3x)]/x = 3 · lim(u→0) sin(u)/u = 3 · 1 = 3.

The key insight is recognizing the standard form sin(u)/u and using substitution to apply the fundamental limit.

Question 2

Determine the value of lim(x→2) [x³ − 8]/(x − 2).

Accepted Answer

Setup: This limit involves a rational function with a removable discontinuity. The numerator and denominator both approach 0 as x → 2, so we must simplify using algebraic factorization (difference of cubes).

Solution:
Step 1: Recognize that direct substitution gives 0/0 (indeterminate form).
Step 2: Factor the numerator using the difference of cubes formula: a³ − b³ = (a − b)(a² + ab + b²).
Step 3: Apply with a = x and b = 2: x³ − 8 = (x − 2)(x² + 2x + 4).
Step 4: Rewrite the limit: lim(x→2) [(x − 2)(x² + 2x + 4)]/(x − 2).
Step 5: Cancel the common factor (x − 2) for x ≠ 2: lim(x→2) (x² + 2x + 4).
Step 6: Apply direct substitution: (2)² + 2(2) + 4 = 4 + 4 + 4 = 12.

The limit equals 12.

Question 3

Evaluate lim(x→0) [sin(3x) / x].

Accepted Answer

This is a standard limit that requires the fundamental trigonometric limit: lim(u→0) (sin u / u) = 1. We rewrite the given expression to match this form. lim(x→0) [sin(3x) / x] = lim(x→0) [sin(3x) / (3x) · 3] = 3 · lim(x→0) [sin(3x) / (3x)]. Let u = 3x. As x → 0, u → 0. Therefore, lim(x→0) [sin(3x) / (3x)] = lim(u→0) (sin u / u) = 1. Thus, lim(x→0) [sin(3x) / x] = 3 · 1 = 3. The key technique is recognizing when to apply the standard limit formula by introducing a multiplicative constant.

Question 4

Let f(x) = |x − 3|. At which point is f(x) discontinuous?

Accepted Answer

Setup: This question tests understanding of continuity of absolute value functions. A function is continuous at a point if the left-hand limit, right-hand limit, and function value all exist and are equal.

Solution:
Step 1: Identify where the expression inside the absolute value changes sign: x − 3 = 0 ⟹ x = 3.
Step 2: Analyze the function piecewise:
  • For x < 3: f(x) = −(x − 3) = 3 − x
  • For x ≥ 3: f(x) = x − 3
Step 3: Check continuity at x = 3:
  • Left-hand limit: lim(x→3⁻) f(x) = lim(x→3⁻) (3 − x) = 3 − 3 = 0
  • Right-hand limit: lim(x→3⁺) f(x) = lim(x→3⁺) (x − 3) = 3 − 3 = 0
  • Function value: f(3) = |3 − 3| = 0
Step 4: Since lim(x→3⁻) f(x) = lim(x→3⁺) f(x) = f(3) = 0, the function IS continuous at x = 3.
Step 5: For all other points, f(x) is a linear function (either 3 − x or x − 3), which is continuous everywhere.

Conclusion: f(x) = |x − 3| is continuous everywhere, including at x = 3.

Question 5

Determine whether f(x) = (x² - 4)/(x - 2) is continuous at x = 2.

Accepted Answer

The function f(x) = (x² - 4)/(x - 2) is not defined at x = 2 because the denominator equals zero. For a function to be continuous at a point, it must be defined at that point. Since f(2) is undefined, f(x) is discontinuous at x = 2. Note: The limit lim(x→2) f(x) exists and equals 4 (by factoring: (x² - 4)/(x - 2) = (x - 2)(x + 2)/(x - 2) = x + 2 for x ≠ 2, so lim(x→2) f(x) = 4). However, the existence of a limit does not guarantee continuity; the function must also be defined at the point. This is a removable discontinuity.

Question 6

Find lim(x→∞) [(2x² + 3x + 1) / (5x² - 2x + 4)].

Accepted Answer

For limits of rational functions as x → ∞, we compare the degrees of the numerator and denominator polynomials. Here, both have degree 2. When degrees are equal, the limit is the ratio of the leading coefficients. Numerator: 2x² (leading term with coefficient 2). Denominator: 5x² (leading term with coefficient 5). Therefore, lim(x→∞) [(2x² + 3x + 1) / (5x² - 2x + 4)] = 2/5. Alternatively, divide numerator and denominator by x²: [(2 + 3/x + 1/x²) / (5 - 2/x + 4/x²)]. As x → ∞, all terms with x in the denominator vanish: (2 + 0 + 0) / (5 - 0 + 0) = 2/5.

Question 7

Let f(x) = |x - 2|. At which point is f(x) discontinuous?

Accepted Answer

The function f(x) = |x - 2| is defined as: f(x) = x - 2 when x ≥ 2, and f(x) = -(x - 2) = 2 - x when x < 2. To check continuity at x = 2, we verify three conditions: (1) f(2) is defined: f(2) = |2 - 2| = 0. ✓ (2) lim(x→2⁻) f(x) = lim(x→2⁻) (2 - x) = 2 - 2 = 0. ✓ (3) lim(x→2⁺) f(x) = lim(x→2⁺) (x - 2) = 2 - 2 = 0. ✓ All three conditions are satisfied, so f is continuous at x = 2. For all other points, f is a linear function (either y = x - 2 or y = 2 - x), which is continuous everywhere. Therefore, f(x) is continuous at every point in ℝ.

Question 8

Evaluate lim(x→∞) [5x² + 3x + 2]/(2x² − x + 1).

Accepted Answer

Setup: This limit involves a rational function where both numerator and denominator are polynomials of the same degree. We must divide both by the highest power of x to evaluate the limit at infinity.

Solution:
Step 1: Identify the highest power of x in both numerator and denominator: x².
Step 2: Divide both numerator and denominator by x²:
  lim(x→∞) [5x² + 3x + 2]/(2x² − x + 1) = lim(x→∞) [(5x²/x²) + (3x/x²) + (2/x²)]/[(2x²/x²) − (x/x²) + (1/x²)]
Step 3: Simplify each term:
  = lim(x→∞) [5 + (3/x) + (2/x²)]/[2 − (1/x) + (1/x²)]
Step 4: As x → ∞, all terms with x in the denominator approach 0:
  = [5 + 0 + 0]/[2 − 0 + 0] = 5/2

The limit equals 5/2.

Question 9

Determine the value of lim(x→2) [(x³ − 8) / (x − 2)].

Accepted Answer

Direct substitution at x = 2 gives (8 − 8)/(2 − 2) = 0/0, an indeterminate form. We must simplify algebraically. Recognize that x³ − 8 is a difference of cubes: x³ − 8 = x³ − 2³ = (x − 2)(x² + 2x + 4). Therefore, [(x³ − 8)/(x − 2)] = [(x − 2)(x² + 2x + 4)]/(x − 2) = x² + 2x + 4 for x ≠ 2. Now apply the limit: lim(x→2) (x² + 2x + 4) = 4 + 4 + 4 = 12. The key is recognizing the difference of cubes factorization and canceling the common factor.

Question 10

Let f(x) = |x − 3|. Determine whether f is continuous at x = 3.

Accepted Answer

To determine continuity at x = 3, we check three conditions: (1) f(3) is defined, (2) lim(x→3) f(x) exists, and (3) lim(x→3) f(x) = f(3). First, f(3) = |3 − 3| = 0, so the function is defined. Next, we compute the left and right limits. For x < 3, f(x) = −(x − 3) = 3 − x, so lim(x→3−) f(x) = 3 − 3 = 0. For x > 3, f(x) = x − 3, so lim(x→3+) f(x) = 3 − 3 = 0. Since both one-sided limits equal 0, lim(x→3) f(x) = 0. Finally, lim(x→3) f(x) = 0 = f(3), so all three conditions are satisfied. Therefore, f is continuous at x = 3. The absolute value function is continuous everywhere, including at points where the expression inside changes sign.

Question 11

Evaluate lim(x→∞) [(2x² + 5x + 1) / (3x² − 4x + 2)].

Accepted Answer

For limits of rational functions as x → ∞, we compare the degrees of the numerator and denominator polynomials. Both the numerator (2x² + 5x + 1) and denominator (3x² − 4x + 2) have degree 2. When degrees are equal, the limit is the ratio of the leading coefficients. The leading coefficient of the numerator is 2, and the leading coefficient of the denominator is 3. Therefore, lim(x→∞) [(2x² + 5x + 1) / (3x² − 4x + 2)] = 2/3. Alternatively, we can divide both numerator and denominator by x² (the highest power): lim(x→∞) [(2 + 5/x + 1/x²) / (3 − 4/x + 2/x²)] = (2 + 0 + 0) / (3 − 0 + 0) = 2/3.

Question 12

Let f(x) = (x² + 3x + 2)/(x + 1) for x ≠ -1. Find lim(x→-1) f(x).

Accepted Answer

To find lim(x→-1) f(x), we first attempt direct substitution. Substituting x = -1 gives (1 - 3 + 2)/(0) = 0/0, an indeterminate form. We must simplify the rational function. Factor the numerator: x² + 3x + 2 = (x + 1)(x + 2). Therefore, f(x) = (x + 1)(x + 2)/(x + 1) = x + 2 for x ≠ -1. Now, lim(x→-1) f(x) = lim(x→-1) (x + 2) = -1 + 2 = 1. The limit exists and equals 1, even though f(-1) is undefined.

Question 13

Determine the value of lim(x→0) [sin(3x) / (5x)].

Accepted Answer

We use the standard limit identity: lim(u→0) (sin u / u) = 1. Rewrite the expression: sin(3x)/(5x) = (sin(3x)/(3x)) · (3x)/(5x) = (sin(3x)/(3x)) · (3/5). As x → 0, let u = 3x, so u → 0. Then lim(x→0) [sin(3x)/(3x)] = lim(u→0) (sin u / u) = 1. Therefore, lim(x→0) [sin(3x)/(5x)] = 1 · (3/5) = 3/5.

Question 14

Let f(x) = (x² + 3x + 2)/(x + 1) for x ≠ −1. Find lim(x→−1) f(x).

Accepted Answer

Setup: We need to evaluate the limit of a rational function at a point where direct substitution gives 0/0 (indeterminate form). This tests algebraic simplification before limit evaluation.

Solution:
Step 1: Factor the numerator: x² + 3x + 2 = (x + 1)(x + 2).
Step 2: Rewrite f(x) = [(x + 1)(x + 2)]/(x + 1).
Step 3: For x ≠ −1, cancel the common factor: f(x) = x + 2.
Step 4: Now apply direct substitution: lim(x→−1) f(x) = lim(x→−1) (x + 2) = −1 + 2 = 1.

The limit exists and equals 1 because the indeterminate form was resolved by factorization and cancellation.

Question 15

If f(x) = (x³ - 8)/(x - 2) for x ≠ 2, then lim(x→2) f(x) equals:

Accepted Answer

We need to find lim(x→2) (x³ - 8)/(x - 2). First, factor the numerator using the difference of cubes formula: a³ - b³ = (a - b)(a² + ab + b²). Here, x³ - 8 = x³ - 2³ = (x - 2)(x² + 2x + 4). Therefore, for x ≠ 2: f(x) = (x - 2)(x² + 2x + 4)/(x - 2) = x² + 2x + 4. Now, lim(x→2) f(x) = lim(x→2) (x² + 2x + 4) = 4 + 4 + 4 = 12.

Question 16

Let f(x) = |x² - 4| for x ∈ ℝ. At which of the following points is f(x) NOT continuous?

Accepted Answer

To determine continuity of f(x) = |x² - 4|, we analyze the function inside the absolute value.

Step 1: Identify critical points where the expression inside absolute value changes sign.
x² - 4 = 0 ⟹ x = ±2

Step 2: Rewrite f(x) piecewise:
f(x) = { x² - 4,    if x² ≥ 4 (i.e., x ≤ -2 or x ≥ 2)
       { -(x² - 4), if x² < 4 (i.e., -2 < x < 2)

Step 3: Check continuity at x = 2:
- Left limit: lim(x→2⁻) f(x) = lim(x→2⁻) (4 - x²) = 4 - 4 = 0
- Right limit: lim(x→2⁺) f(x) = lim(x→2⁺) (x² - 4) = 4 - 4 = 0
- f(2) = |4 - 4| = 0
Left limit = Right limit = f(2), so f is continuous at x = 2.

Step 4: Check continuity at x = -2:
- Left limit: lim(x→-2⁻) f(x) = lim(x→-2⁻) (x² - 4) = 4 - 4 = 0
- Right limit: lim(x→-2⁺) f(x) = lim(x→-2⁺) (4 - x²) = 4 - 4 = 0
- f(-2) = |4 - 4| = 0
Left limit = Right limit = f(-2), so f is continuous at x = -2.

Step 5: For all other points, f(x) is a composition of continuous functions (polynomial and absolute value), hence continuous everywhere.

Conclusion: f(x) = |x² - 4| is continuous everywhere on ℝ, including at x = ±2. The absolute value function smooths out the corner algebraically.

Question 17

Let f(x) = (x² + 1)/(x - 1). At which point(s) is f(x) discontinuous, and what is the nature of the discontinuity?

Accepted Answer

A rational function is discontinuous where its denominator equals zero. Setting x - 1 = 0 gives x = 1. At x = 1, the numerator is 1² + 1 = 2 ≠ 0, so the denominator is zero while the numerator is non-zero. This creates a vertical asymptote, which is an infinite discontinuity (not removable). The function is continuous everywhere else on ℝ.

Question 18

Evaluate lim(x→0) [sin(3x) - 3sin(x)] / x³.

Accepted Answer

This limit requires L'Hôpital's rule or Taylor series. Using Taylor expansion: sin(3x) = 3x - (3x)³/6 + O(x⁵) = 3x - 9x³/2 + O(x⁵) and sin(x) = x - x³/6 + O(x⁵). Therefore: sin(3x) - 3sin(x) = [3x - 9x³/2 + ...] - 3[x - x³/6 + ...] = 3x - 9x³/2 - 3x + x³/2 + ... = -4x³ + O(x⁵). Thus: [sin(3x) - 3sin(x)]/x³ = -4x³/x³ + O(x²) = -4 as x→0. Alternatively, using L'Hôpital's rule three times (since 0/0 form persists through two applications) yields -4.

Question 19

For what value of k is the function f(x) = {(x² - 4)/(x - 2) if x ≠ 2; k if x = 2} continuous at x = 2?

Accepted Answer

For f to be continuous at x = 2, we need lim(x→2) f(x) = f(2) = k. For x ≠ 2, f(x) = (x² - 4)/(x - 2) = (x - 2)(x + 2)/(x - 2) = x + 2 (after cancellation). Therefore, lim(x→2) f(x) = lim(x→2) (x + 2) = 4. For continuity, k = 4.

Question 20

Let f(x) = |x - 1|. Which of the following statements about f is correct?

Accepted Answer

The function f(x) = |x - 1| is defined for all x ∈ ℝ. We can write f(x) = {x - 1 if x ≥ 1; 1 - x if x < 1}. To check continuity at x = 1: lim(x→1⁺) f(x) = lim(x→1⁺) (x - 1) = 0, lim(x→1⁻) f(x) = lim(x→1⁻) (1 - x) = 0, and f(1) = |1 - 1| = 0. Since all three are equal, f is continuous at x = 1 and everywhere else. To check differentiability at x = 1: f'(1⁺) = 1 and f'(1⁻) = -1. Since the left and right derivatives are unequal, f is not differentiable at x = 1. The function is continuous everywhere but not differentiable at x = 1.

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