27-02-2025 C | Engineering Mathematics | GATE & ESE

Key Concepts

• Standard Limits: Predefined limit formulas used for solving limit problems.
• Indeterminate Forms: Expressions like 0/0 or ∞/∞ where the limit cannot be directly determined.
• L'Hôpital's Rule: A method for evaluating indeterminate forms by differentiating the numerator and denominator.
• Continuity: A function is continuous at a point if the limit exists at that point and equals the function's value.
• Differentiability: A function is differentiable at a point if its derivative exists at that point.
• Left-Hand Limit (LHL) & Right-Hand Limit (RHL): The limit of a function as x approaches a value from the left and right, respectively.
• Left-Hand Derivative (LHD) & Right-Hand Derivative (RHD): The derivative of a function as x approaches a value from the left and right, respectively.
• Corner Points: Points on a graph where the function is continuous but not differentiable due to a sharp edge.

Standard Limits and Their Application

The video begins by listing standard limits, emphasizing their importance as formulas for solving limit problems. These limits include:

• lim (x→a) f(x) = L exists if and only if LHL = RHL.
• Examples of standard limits are referenced throughout the problem-solving section.

Indeterminate Forms and L'Hôpital's Rule

The video explains that indeterminate forms (0/0, ∞/∞) arise when direct substitution yields an undefined result.

• Example: In question 12, substituting x=0 into (e^(4x) - 1) / sin(2x) results in 0/0.
• L'Hôpital's Rule: To resolve indeterminate forms, the video demonstrates L'Hôpital's Rule:
  • Differentiate the numerator and denominator separately.
  • Re-evaluate the limit.
  • Example: Applying L'Hôpital's Rule to (e^(4x) - 1) / sin(2x) yields (4e^(4x)) / (2cos(2x)), which evaluates to 2 as x approaches 0.

Question 12: Applying L'Hôpital's Rule

The video solves the limit problem: lim (x→0) (e^(4x) - 1) / sin(2x).

1. Identify Indeterminate Form: Direct substitution yields 0/0.
2. Apply L'Hôpital's Rule: Differentiate numerator and denominator:
   • d/dx (e^(4x) - 1) = 4e^(4x)
   • d/dx (sin(2x)) = 2cos(2x)
3. Re-evaluate Limit: lim (x→0) (4e^(4x)) / (2cos(2x)) = (4 * 1) / (2 * 1) = 2.

Question 13: Using Standard Limits and L'Hôpital's Rule

The video tackles lim (x→0) (x - sin(x)) / (x - tan(x)).

1. Identify Indeterminate Form: Direct substitution yields 0/0.
2. Apply L'Hôpital's Rule: Differentiate numerator and denominator:
   • d/dx (x - sin(x)) = 1 - cos(x)
   • d/dx (x - tan(x)) = 1 - sec²(x)
3. Re-evaluate Limit: lim (x→0) (1 - cos(x)) / (1 - sec²(x)).
4. Further Simplification: The expression is manipulated using trigonometric identities (sec²(x) - 1 = tan²(x)) and then multiplied and divided by x² to leverage standard limits.
5. Apply Standard Limits:
   • lim (x→0) (1 - cos(x)) / x² = 1/2
   • lim (x→0) tan²(x) / x² = 1
6. Final Result: The limit evaluates to (1/2) / (-1) = -1/2.

Question 14: Limits at Infinity and Polynomials

The video addresses lim (n→∞) (n * Σn²) / Σn³.

1. Summation Formulas: The video uses the formulas:
   • Σn² = n(n+1)(2n+1) / 6
   • Σn³ = (n(n+1) / 2)² = n²(n+1)² / 4
2. Substitute and Simplify: The expression becomes lim (n→∞) (n * n(n+1)(2n+1) / 6) / (n²(n+1)² / 4).
3. Identify Highest Degree Terms: In both numerator and denominator, the highest degree term is n².
4. Divide by Highest Degree: Divide both numerator and denominator by n².
5. Evaluate Limit: As n approaches infinity, terms like 1/n approach 0, and the limit simplifies to 4/3.

Question 15: Rationalization and Limits at Infinity

The video solves lim (x→∞) √(x² - 2x + 2) - x.

1. Identify Indeterminate Form: Direct substitution yields ∞ - ∞.
2. Rationalization: Multiply and divide by the conjugate √(x² - 2x + 2) + x.
3. Simplify: The expression becomes lim (x→∞) (x² - 2x + 2 - x²) / (√(x² - 2x + 2) + x) = lim (x→∞) (-2x + 2) / (√(x² - 2x + 2) + x).
4. Divide by x: Divide both numerator and denominator by x.
5. Evaluate Limit: As x approaches infinity, terms like 2/x approach 0, and the limit simplifies to -2 / (1 + 1) = -1.

Question 5: L'Hôpital's Rule and Trigonometric Functions

The video evaluates lim (x→π/2) (1 - sin³(x)) / cos²(x).

1. Identify Indeterminate Form: Direct substitution yields 0/0.
2. Apply L'Hôpital's Rule: Differentiate numerator and denominator:
   • d/dx (1 - sin³(x)) = -3sin²(x)cos(x)
   • d/dx (cos²(x)) = -2cos(x)sin(x)
3. Re-evaluate Limit: lim (x→π/2) (-3sin²(x)cos(x)) / (-2cos(x)sin(x)).
4. Simplify: Cancel out common terms (cos(x), sin(x)).
5. Final Result: The limit evaluates to 3sin(π/2) / 2 = 3/2.

Question 20: Recognizing Standard Limits

The video addresses lim (α→0) (x^α - 1) / α.

1. Recognize Standard Limit: The expression resembles the standard limit lim (x→0) (a^x - 1) / x = ln(a).
2. Apply Standard Limit: In this case, 'x' is constant, and 'α' is the variable approaching 0. Therefore, the limit is ln(x).

Question 22: Substitution and Standard Limits

The video solves lim (x→∞) (e^(1/5x) - 1) * (5x + x) / (5x * sin(1/x)).

1. Substitution: Let y = 1/x. As x approaches infinity, y approaches 0.
2. Rewrite Limit: The limit becomes lim (y→0) (e^(y/5) - 1) * (5/y + 1/y) / (5/y * sin(y)).
3. Simplify: The expression simplifies to lim (y→0) (e^(y/5) - 1) * (6/y) / (5/y * sin(y)) = lim (y→0) (e^(y/5) - 1) * 6 / (5 * sin(y)).
4. Apply Standard Limits:
   • lim (y→0) (e^(y/5) - 1) / y = 1/5
   • lim (y→0) sin(y) / y = 1
5. Final Result: The limit evaluates to (1/5) * 6 / (5 * 1) = 6/25.

Question 7: 1^∞ Form and Standard Limits

The video evaluates lim (x→0) (tan(x) / x)^(1/x²).

1. Identify Indeterminate Form: Direct substitution yields 1^∞.
2. Apply Standard Limit: Use the formula lim (x→a) f(x)^g(x) = e^(lim (x→a) g(x) * (f(x) - 1)) when the limit is of the form 1^∞.
3. Rewrite Limit: The limit becomes e^(lim (x→0) (1/x²) * (tan(x) / x - 1)).
4. Simplify: The expression inside the exponent becomes lim (x→0) (tan(x) - x) / x³.
5. Apply L'Hôpital's Rule: Differentiate numerator and denominator:
   • d/dx (tan(x) - x) = sec²(x) - 1 = tan²(x)
   • d/dx (x³) = 3x²
6. Re-evaluate Limit: The limit becomes lim (x→0) tan²(x) / (3x²).
7. Apply Standard Limit: lim (x→0) tan²(x) / x² = 1.
8. Final Result: The limit evaluates to e^(1/3).

Question 10: Limits with Absolute Values

The video solves lim (x→0+) x / |x|.

1. Definition of Absolute Value: |x| = x for x > 0 and |x| = -x for x < 0.
2. Apply Definition: Since x approaches 0 from the positive side (0+), x > 0, so |x| = x.
3. Simplify: The expression becomes lim (x→0+) x / x = 1.

Question 8: Finding Unknowns with Finite Limits

The video addresses lim (x→0) (sin(2x) + a sin(x)) / x³ = finite. Find 'a' and the limit.

1. Identify Indeterminate Form: Direct substitution yields 0/0.
2. Apply L'Hôpital's Rule: Differentiate numerator and denominator:
   • d/dx (sin(2x) + a sin(x)) = 2cos(2x) + a cos(x)
   • d/dx (x³) = 3x²
3. Re-evaluate Limit: lim (x→0) (2cos(2x) + a cos(x)) / (3x²).
4. Condition for Finite Limit: For the limit to be finite, the numerator must also approach 0 as x approaches 0. This implies 2 + a = 0, so a = -2.
5. Substitute 'a': The limit becomes lim (x→0) (2cos(2x) - 2cos(x)) / (3x²).
6. Apply L'Hôpital's Rule Again: Differentiate numerator and denominator:
   • d/dx (2cos(2x) - 2cos(x)) = -4sin(2x) + 2sin(x)
   • d/dx (3x²) = 6x
7. Re-evaluate Limit: lim (x→0) (-4sin(2x) + 2sin(x)) / (6x).
8. Apply Standard Limits: Use lim (x→0) sin(x) / x = 1 and lim (x→0) sin(2x) / x = 2.
9. Final Result: The limit evaluates to (-4 * 2 + 2 * 1) / 6 = -6/6 = -1.

Continuity: Definition and Examples

The video defines continuity: a function f(x) is continuous at x=a if lim (x→a) f(x) = f(a). This means the left-hand limit, right-hand limit, and the function's value at 'a' must all be equal.

• Graphical Explanation: The video uses a graphical representation to illustrate the difference between a function having a limit and being continuous. For a function to be continuous, the graph must be connected at that point.

Question 9: Applying Continuity Definition

The video solves for λ if f(x) is continuous at x=3, where f(x) = x + λ for x < 3, 3x - 5 for x > 3, and 4 at x = 3.

1. Apply Continuity Definition: f(3-) = f(3+) = f(3).
2. Evaluate Limits:
   • f(3-) = lim (x→3-) (x + λ) = 3 + λ
   • f(3+) = lim (x→3+) (3x - 5) = 4
   • f(3) = 4
3. Solve for λ: 3 + λ = 4, so λ = 1.

Question 8: Continuity and Limits

The video finds 'a' if f(x) is continuous at x=3, where f(x) = (x² - (a+3)x + 3a) / (x - 3) for x ≠ 3 and 2a - 3 at x = 3.

1. Apply Continuity Definition: lim (x→3) f(x) = f(3).
2. Evaluate Limit: lim (x→3) (x² - (a+3)x + 3a) / (x - 3).
3. Identify Indeterminate Form: Direct substitution yields 0/0.
4. Apply L'Hôpital's Rule: Differentiate numerator and denominator:
   • d/dx (x² - (a+3)x + 3a) = 2x - (a+3)
   • d/dx (x - 3) = 1
5. Re-evaluate Limit: lim (x→3) (2x - (a+3)) / 1 = 6 - (a+3) = 3 - a.
6. Equate to Function Value: 3 - a = 2a - 3.
7. Solve for 'a': 3a = 6, so a = 2.

Question 1: Continuity and Piecewise Functions

The video finds A and B for which f(x) is continuous everywhere, where f(x) is a piecewise function defined differently for x < 1, 1 < x < 3, and x > 3.

1. Apply Continuity at x=1: f(1-) = f(1+).
   • lim (x→1-) (2x + 1) = 3
   • lim (x→1+) (Ax² + B) = A + B
   • Therefore, A + B = 3.
2. Apply Continuity at x=3: f(3-) = f(3+).
   • lim (x→3-) (Ax² + B) = 9A + B
   • lim (x→3+) (5x + 2A) = 15 + 2A
   • Therefore, 9A + B = 15 + 2A, which simplifies to 7A + B = 15.
3. Solve the System of Equations: Solve the system A + B = 3 and 7A + B = 15 to find A = 2 and B = 1.

Differentiability: Definition and Properties

The video defines differentiability: a function f(x) is differentiable at x=a if the limit of (f(x) - f(a)) / (x - a) exists as x approaches a. This is equivalent to the left-hand derivative (LHD) equaling the right-hand derivative (RHD).

• Rate of Change: Differentiability is explained as the rate of change of y with respect to x (dy/dx).
• Smooth Curves: A smooth curve is differentiable.
• Corner Points: A function is not differentiable at corner points (sharp edges or sudden changes).
• Continuity and Differentiability: Every differentiable function is continuous, but the converse is not always true.
• Example: f(x) = |x| is continuous everywhere but not differentiable at x = 0 (corner point).

Question 3: Differentiability with Absolute Values

The video finds f'(π/4) for f(x) = sin(|x|).

1. Rewrite without Absolute Value:
   • f(x) = -sin(x) for x < 0
   • f(x) = sin(x) for x ≥ 0
2. Determine Relevant Function: Since π/4 > 0, use f(x) = sin(x).
3. Differentiate: f'(x) = cos(x).
4. Evaluate at x=π/4: f'(π/4) = cos(π/4) = 1/√2.

Question 11: Differentiability and Piecewise Functions

The video determines α and β if f(x) is differentiable at x=1, where f(x) = 2x + 1 for x ≤ 1 and αx² + βx for x > 1.

1. Apply Differentiability Condition: f'(1-) = f'(1+).
   • f'(x) = 2 for x < 1, so f'(1-) = 2.
   • f'(x) = 2αx + β for x > 1, so f'(1+) = 2α + β.
   • Therefore, 2α + β = 2.
2. Apply Continuity Condition: f(1-) = f(1+).
   • lim (x→1-) (2x + 1) = 3
   • lim (x→1+) (αx² + βx) = α + β
   • Therefore, α + β = 3.
3. Solve the System of Equations: Solve the system 2α + β = 2 and α + β = 3 to find α = -1 and β = 4.

Conclusion

The video provides a comprehensive overview of limits, continuity, and differentiability, including definitions, properties, and problem-solving techniques. It emphasizes the importance of understanding standard limits, L'Hôpital's Rule, and the relationship between continuity and differentiability. The video also highlights the use of graphical interpretations and the application of these concepts to piecewise functions and functions involving absolute values.