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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 2.3.49
Chapter 2, Problem 2.3.49

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


lim x→0 x sin (1/x) = 0


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Understand the formal definition of a limit: For the limit \( \lim_{x \to a} f(x) = L \), for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon \).
Identify the function and limit in the problem: Here, \( f(x) = x \sin(1/x) \) and we want to prove \( \lim_{x \to 0} x \sin(1/x) = 0 \).
Express \( |f(x) - L| \) using the given function and limit: Since \( L = 0 \), we have \( |x \sin(1/x) - 0| = |x \sin(1/x)| \).
Use the property of the sine function: Since \( |\sin(1/x)| \leq 1 \) for all \( x \neq 0 \), it follows that \( |x \sin(1/x)| \leq |x| \).
Choose \( \delta = \epsilon \) to satisfy the limit definition: For \( |x| < \delta \), we have \( |x \sin(1/x)| \leq |x| < \epsilon \), thus proving the limit statement.

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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 a is L if, for every ε > 0, there exists a δ > 0 such that whenever 0 < |x - a| < 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 in some interval around a (except possibly at a), and if the limits of f(x) and h(x) as x approaches a are both L, then the limit of g(x) as x approaches a is also L. This theorem is particularly useful for functions like x sin(1/x).
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Behavior of Oscillating Functions

The function sin(1/x) oscillates between -1 and 1 as x approaches 0, which means it does not settle at a single value. However, when multiplied by x, which approaches 0, the product x sin(1/x) is squeezed to 0. Understanding the behavior of oscillating functions is essential for analyzing limits involving such terms.
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Related Practice
Textbook Question

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limh→0− h / sin 3h

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Textbook Question

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 → ±∞ f(x) = 0, lim x → 2⁻ f(x) = ∞, and lim x → 2⁺ f(x) = ∞

Textbook Question

Horizontal and Vertical Asymptotes


Determine the domain and range of y = (√16―x²) / (x―2).

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Textbook Question

Limits of Rational Functions


In Exercises 13–22, find the limit of each rational function (a) as x → ∞ and (b) as x → −∞. Write ∞ or −∞ where appropriate.


f(x) = (x + 1)/(x² + 3)

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