Textbook Question
Limits and Infinity
Find the limits in Exercises 37–46.
2x² + 3
lim -------------
x→⁻∞ 5x² + 7
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 2.8.84
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))
Verified step by step guidance1
Identify the expression whose limit you need to find: \( \lim_{x \to \infty} (\sqrt{x + 9} - \sqrt{x + 4}) \).
To simplify the expression, multiply and divide by the conjugate: \( \frac{(\sqrt{x + 9} - \sqrt{x + 4})(\sqrt{x + 9} + \sqrt{x + 4})}{\sqrt{x + 9} + \sqrt{x + 4}} \).
The numerator becomes a difference of squares: \((x + 9) - (x + 4) = 5\).
The expression simplifies to \( \frac{5}{\sqrt{x + 9} + \sqrt{x + 4}} \).
As \( x \to \infty \), both \( \sqrt{x + 9} \) and \( \sqrt{x + 4} \) approach \( \sqrt{x} \), so the denominator approaches \( 2\sqrt{x} \). Thus, the limit becomes \( \lim_{x \to \infty} \frac{5}{2\sqrt{x}} \), which approaches 0.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits
Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. In this context, we are interested in the behavior of the function as x approaches infinity, which helps us understand the end behavior of the function and how it simplifies under such conditions.
Recommended video:
05:50
One-Sided Limits
Conjugates
The conjugate of a binomial expression is formed by changing the sign between two terms. In the context of limits, multiplying by the conjugate can help eliminate square roots or simplify expressions, making it easier to evaluate the limit. This technique is particularly useful when dealing with differences of square roots.
Recommended video:
06:13Limits of Rational Functions with Radicals
Rationalization
Rationalization is a method used to eliminate radicals from the denominator or simplify expressions involving square roots. By multiplying the numerator and denominator by the conjugate, we can transform the expression into a more manageable form, allowing for easier calculation of limits, especially as x approaches infinity.
Recommended video:
6:04Intro to Rational Functions
Related Practice
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
Using the Formal Definition
Prove the limit statements in Exercises 37–50.
lim x→0 x² sin (1/x) = 0
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Textbook Question
At what points are the functions in Exercises 13–30 continuous?
y = √(2x + 3)