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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 2 - Limits and ContinuityProblem 2.3.50
Chapter 2, Problem 2.3.50

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


lim x→0 x² sin (1/x) = 0


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Understand the formal definition of a limit: For a function f(x) to have a limit L as x approaches a, for every ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε.
Identify the function and the limit: Here, f(x) = x² sin(1/x) and we want to prove that lim x→0 f(x) = 0.
Set up the inequality using the formal definition: We need to show that for every ε > 0, there exists a δ > 0 such that if 0 < |x| < δ, then |x² sin(1/x) - 0| < ε.
Simplify the expression: Notice that |x² sin(1/x)| = |x²| |sin(1/x)|. Since |sin(1/x)| ≤ 1 for all x, we have |x² sin(1/x)| ≤ |x²|.
Choose δ: To ensure |x²| < ε, we can choose δ = √ε. Then, for 0 < |x| < δ, we have |x²| < δ² = ε, which satisfies the condition |x² sin(1/x)| < ε, proving the limit.

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Key Concepts

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

Limit Definition

The formal definition of a limit states that for a function f(x), the limit as x approaches a value c is L if, for every ε > 0, there exists a δ > 0 such that whenever 0 < |x - c| < δ, it follows that |f(x) - L| < ε. This definition is crucial for proving limit statements rigorously.
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Squeeze Theorem

The Squeeze Theorem is a method used to find limits of functions that are difficult to evaluate directly. It states that if f(x) ≤ g(x) ≤ h(x) for all x near c (except possibly at c), and if the limits of f(x) and h(x) as x approaches c are both L, then the limit of g(x) as x approaches c is also L. This theorem is particularly useful for functions involving oscillatory behavior, like sin(1/x).
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Behavior of Oscillatory Functions

Oscillatory functions, such as sin(1/x), do not settle at a single value as x approaches 0; instead, they oscillate between -1 and 1. Understanding this behavior is essential when evaluating limits involving such functions, as it allows us to analyze their impact on the overall limit, particularly when multiplied by terms that approach zero, like x² in this case.
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Related Practice
Textbook Question

Finding Limits of Differences When x → ±∞


Find the limits in Exercises 84–90. (Hint: Try multiplying and dividing by the conjugate.)


lim x → ∞ (√(x + 9) − √(x + 4))

Textbook Question

Using the Sandwich Theorem


a. It can be shown that the inequalities 1 − x²/ 6 < (x sin x) / (2−2cos x) < 1 hold for all values of x close to zero (except for x = 0). What, if anything, does this tell you about limx→0 (x sin x) / (2 − 2cos x)?


Give reasons for your answer.


[Technology Exercise] b. Graph y = 1 − (x²/6), y=(x sinx)/(2 − 2cos x), and y = 1 together for −2 ≤ x ≤2. Comment on the behavior of the graphs as x→0.

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 − 15x + 1 = 0 (three roots)

Textbook Question

Domains and Asymptotes


Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.


y = 2x / (x² − 1)

Textbook Question

If functions f(x) and g(x) are continuous for 0 ≤ x ≤ 1, could f(x)/g(x) possibly be discontinuous at a point of [0,1]? Give reasons for your answer.

Textbook Question

Finding Limits


In Exercises 25–28, find the limit of g(x) as x approaches the indicated value.



lim (4g(x))¹/³ = 2

x →0