Textbook Question
Limits of quotients
Find the limits in Exercises 23–42.
limu→1 (u⁴ − 1)/(u³ − 1)
Ch. 2 - Limits and Continuity
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 2, Problem 2.3.45
Using the Formal Definition
Prove the limit statements in Exercises 37–50.
limx→−3 (x² − 9) / (x + 3) = −6
Verified step by step guidance1
Understand the problem: We need to prove the limit \( \lim_{x \to -3} \frac{x^2 - 9}{x + 3} = -6 \) using the formal definition of a limit.
Simplify the expression: Notice that \( x^2 - 9 \) can be factored as \((x - 3)(x + 3)\). This allows us to rewrite the expression as \( \frac{(x - 3)(x + 3)}{x + 3} \). For \( x \neq -3 \), this simplifies to \( x - 3 \).
Apply the limit: Now, we need to find \( \lim_{x \to -3} (x - 3) \). As \( x \) approaches \(-3\), the expression \( x - 3 \) approaches \(-3 - 3 = -6\).
Use the formal definition of a limit: For every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x + 3| < \delta \), then \( |(x - 3) + 6| < \epsilon \).
Verify the condition: Simplifying \( |(x - 3) + 6| = |x + 3| \), we can choose \( \delta = \epsilon \). Therefore, for any \( \epsilon > 0 \), if \( 0 < |x + 3| < \epsilon \), then \( |x + 3| < \epsilon \), proving the limit is \(-6\).
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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 rigorously proving limit statements.
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Definition of the Definite Integral
Factoring and Simplifying Expressions
To evaluate limits, especially when direct substitution results in an indeterminate form like 0/0, factoring and simplifying the expression can help. In the given limit, the expression (x² - 9) can be factored into (x - 3)(x + 3), allowing for cancellation of the (x + 3) term.
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Substitution in Limits
Once the expression is simplified, substitution can be used to find the limit. In this case, after canceling the common factor, substituting x = -3 into the simplified expression yields the limit value. Understanding how to apply substitution correctly is essential for limit evaluation.
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Related Practice
Textbook Question
At what points are the functions in Exercises 13–30 continuous?
y = (2x – 1)¹/³
Textbook Question
Using limθ→0 sin θ / θ = 1
Find the limits in Exercises 23–46.
limx→0 (x² − x + sin x) / 2x
Textbook Question
At what points are the functions in Exercises 13–30 continuous?
y = √(x⁴ +1)/(1 + sin² x)
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
Suppose that a function f(x) is defined for all x in [-1,1]. Can anything be said about the existence of limx→0 f(x)? Give reasons for your answer.