Ch. 2 - Limits and Continuity

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 2 - Limits and Continuity

Problem 2.6.80

Chapter 2, Problem 2.6.80

In Exercises 77–80, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.)

lim x → ±∞ k(x) = 1, lim x → 1⁻ k(x) = ∞, and lim x → 1⁺ k(x) = −∞

Verified step by step guidance

Identify the type of function that can have different behaviors at different points. A rational function is a good candidate because it can have vertical asymptotes and horizontal asymptotes.

To satisfy lim x → ±∞ k(x) = 1, consider a rational function where the degrees of the numerator and denominator are the same. For example, k(x) = (ax + b)/(cx + d) where a/c = 1.

To satisfy lim x → 1⁻ k(x) = ∞, the function must have a vertical asymptote at x = 1. This can be achieved by having a factor of (x - 1) in the denominator.

To satisfy lim x → 1⁺ k(x) = −∞, the function should approach negative infinity from the right of x = 1. This can be achieved by ensuring the factor (x - 1) in the denominator does not cancel with a similar factor in the numerator.

Combine these insights to construct a function like k(x) = (x + 1)/(x - 1). This function has a horizontal asymptote at y = 1, a vertical asymptote at x = 1, and the behavior around x = 1 matches the given conditions.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits at Infinity

Limits at infinity describe the behavior of a function as the input approaches positive or negative infinity. In this context, lim x → ±∞ k(x) = 1 indicates that as x becomes very large or very small, the function k(x) approaches the value 1. This suggests a horizontal asymptote at y = 1, which is crucial for sketching the graph of the function.

Vertical Asymptotes

A vertical asymptote occurs when a function approaches infinity or negative infinity as the input approaches a specific value. Here, lim x → 1⁻ k(x) = ∞ and lim x → 1⁺ k(x) = −∞ indicate a vertical asymptote at x = 1. This means the function k(x) increases without bound as x approaches 1 from the left and decreases without bound as x approaches 1 from the right.

Piecewise Functions

Piecewise functions are defined by different expressions over different intervals of the domain. In this problem, using a piecewise function can help construct a function that meets the given conditions, such as having different behaviors around x = 1 and as x approaches infinity. This flexibility is essential for creating a function that satisfies all the specified limits.

Recommended video:

05:36

Piecewise Functions

Related Practice

Textbook Question

At what points are the functions in Exercises 13–30 continuous?

y = (2x – 1)¹/³

Textbook Question

Use the Intermediate Value Theorem in Exercises 69–74 to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations.

x³ − 3x − 1 = 0

Textbook Question

In Exercises 1–4, say whether the function graphed is continuous on [−1, 3]. If not, where does it fail to be continuous and why?

<IMAGE>

views

Textbook Question

Suppose limx→c f(x) = 5 and lim x→c g(x) = −2. Find

b. limx→c 2f(x)g(x)

Textbook Question

Using the Sandwich Theorem

a. Suppose that the inequalities 1/2 − x² / 24 < (1 − cos x)/ x² < 1/2 hold for values of x close to zero, except for x = 0 itself. (They do, as you will see in Section 9.9.) What, if anything, does this tell you about limx→0 (1 −cos x)/ x²?

Give reasons for your answer.

[Technology Exercise] b. Graph the equations y=(1/2) − (x²/24), y = (1 - cos x) / x², and y = 1/2 together for −2 ≤ x ≤2. Comment on the behavior of the graphs as x→0.

Textbook Question

Use formal definitions to prove the limit statements in Exercises 93–96.

lim x → −5 (1 / (x + 5)²) = ∞