1.6 Limits And Continuity Homework Answers

1.6 Limits and Continuity: Homework Answers and Comprehensive Guide

This comprehensive guide delves into the intricacies of limits and continuity, providing detailed explanations, solved examples, and strategies to tackle common homework problems in Calculus. We'll explore the fundamental concepts, address potential stumbling blocks, and equip you with the tools to confidently approach any limit and continuity problem.

Understanding Limits

Before tackling homework problems, let's solidify our understanding of limits. A limit describes the behavior of a function as its input approaches a particular value. We write this as:

lim_(x→a) f(x) = L

This means that as x gets arbitrarily close to a, the function f(x) gets arbitrarily close to L. Crucially, the function doesn't need to be defined at x = a for the limit to exist.

Types of Limits

Several types of limits exist, each requiring a slightly different approach:

• One-sided limits: These examine the function's behavior as x approaches a from the left (lim_(x→a⁻) f(x)) or the right (lim_(x→a⁺) f(x)). For a limit to exist, both one-sided limits must exist and be equal.

• Infinite limits: These occur when the function's values approach infinity or negative infinity as x approaches a. We denote these as lim_(x→a) f(x) = ∞ or lim_(x→a) f(x) = -∞.

• Limits at infinity: These describe the function's behavior as x approaches positive or negative infinity (lim_(x→∞) f(x) or lim_(x→-∞) f(x)). These often involve identifying horizontal asymptotes.

Techniques for Evaluating Limits

Several techniques help evaluate limits:

1. Direct Substitution:

The simplest method. If the function is continuous at x = a, simply substitute a into the function: lim_(x→a) f(x) = f(a). However, this doesn't always work (e.g., if there's a division by zero).

2. Algebraic Manipulation:

Often, algebraic simplification (factoring, rationalizing the numerator/denominator, etc.) eliminates indeterminate forms (like 0/0 or ∞/∞) allowing for direct substitution.

Example:

Find lim_(x→2) (x² - 4) / (x - 2)

This limit initially yields 0/0. Factoring the numerator gives:

lim_(x→2) (x - 2)(x + 2) / (x - 2) = lim_(x→2) (x + 2) = 4

3. L'Hôpital's Rule:

Applicable when the limit is in an indeterminate form (0/0 or ∞/∞). L'Hôpital's rule states that if lim_(x→a) f(x)/g(x) is indeterminate, then:

lim_(x→a) f(x)/g(x) = lim_(x→a) f'(x)/g'(x)

(where f'(x) and g'(x) are the derivatives of f(x) and g(x) respectively). Apply this rule repeatedly if necessary until you get a determinate form.

Example:

Find lim_(x→0) sin(x)/x

This limit is of the form 0/0. Applying L'Hôpital's rule:

lim_(x→0) sin(x)/x = lim_(x→0) cos(x)/1 = cos(0) = 1

4. Squeeze Theorem (Sandwich Theorem):

If we can bound a function f(x) between two other functions, g(x) and h(x), such that g(x) ≤ f(x) ≤ h(x) and lim_(x→a) g(x) = lim_(x→a) h(x) = L, then lim_(x→a) f(x) = L.

Understanding Continuity

A function is continuous at a point x = a if:

1. f(a) is defined.
2. lim_(x→a) f(x) exists.
3. lim_(x→a) f(x) = f(a)

If a function is continuous at every point in its domain, it's considered a continuous function.

Types of Discontinuities

Discontinuities arise when a function fails to meet the conditions for continuity:

• Removable Discontinuity: The limit exists, but it doesn't equal the function's value at that point (or the function isn't defined at that point). Often, this can be "fixed" by redefining the function at that point.

• Jump Discontinuity: The one-sided limits exist but are different. There's a "jump" in the function's value.

• Infinite Discontinuity: The function approaches positive or negative infinity as x approaches a. This often corresponds to vertical asymptotes.

Solving Homework Problems: A Step-by-Step Approach

Let's tackle some typical homework problems, illustrating the techniques discussed:

Problem 1: Evaluate lim_(x→3) (x² - 9) / (x - 3)

1. Direct Substitution: Attempting direct substitution yields 0/0, an indeterminate form.

2. Algebraic Manipulation: Factor the numerator: (x² - 9) = (x - 3)(x + 3).

3. Simplification: The expression becomes (x - 3)(x + 3) / (x - 3). We can cancel the (x - 3) terms (since x ≠ 3 as we're approaching 3, not at 3).

4. Evaluation: The limit simplifies to lim_(x→3) (x + 3) = 6.

Problem 2: Evaluate lim_(x→∞) (3x² + 2x - 1) / (x² - 5x + 2)

1. Direct Substitution: Substituting infinity leads to ∞/∞, an indeterminate form.

2. Divide by the highest power of x: Divide both the numerator and denominator by x², the highest power of x in the denominator.

3. Simplification: The expression becomes (3 + 2/x - 1/x²) / (1 - 5/x + 2/x²).

4. Evaluation: As x approaches infinity, the terms 2/x, 1/x², 5/x, and 2/x² all approach zero. Therefore, the limit simplifies to 3/1 = 3.

Problem 3: Determine if the function f(x) = {x² if x < 2; 3x - 2 if x ≥ 2} is continuous at x = 2.

1. Check f(2): f(2) = 3(2) - 2 = 4.

2. Check the left-hand limit: lim_(x→2⁻) f(x) = lim_(x→2⁻) x² = 4.

3. Check the right-hand limit: lim_(x→2⁺) f(x) = lim_(x→2⁺) (3x - 2) = 4.

4. Compare: Since f(2) = lim_(x→2⁻) f(x) = lim_(x→2⁺) f(x) = 4, the function is continuous at x = 2.

Problem 4: Evaluate lim_(x→0) (1 - cos(x)) / x

1. Direct Substitution: This gives 0/0, an indeterminate form.

2. L'Hôpital's Rule: Differentiate the numerator and denominator:

   Numerator derivative: d(1 - cos(x))/dx = sin(x) Denominator derivative: d(x)/dx = 1

3. Simplification: The limit becomes lim_(x→0) sin(x) / 1 = sin(0) = 0.

Problem 5: Using the Squeeze Theorem, find lim_(x→0) x²sin(1/x)

1. Inequality: We know that -1 ≤ sin(1/x) ≤ 1 for all x ≠ 0.

2. Multiplication: Multiplying by x² (which is always non-negative), we get: -x² ≤ x²sin(1/x) ≤ x²

3. Limits: We have lim_(x→0) -x² = 0 and lim_(x→0) x² = 0.

4. Squeeze Theorem: By the Squeeze Theorem, lim_(x→0) x²sin(1/x) = 0.

Conclusion

Mastering limits and continuity is crucial for success in Calculus. This guide provides a strong foundation, combining theoretical understanding with practical problem-solving techniques. Remember to practice regularly, applying these methods to a wide range of problems. By understanding the different types of limits, the various techniques for evaluation, and the criteria for continuity, you'll develop the confidence and skills needed to tackle even the most challenging homework assignments and excel in your calculus studies. Remember to always check your work and consider consulting additional resources if needed. Good luck!