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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 2.38
Chapter 2, Problem 2.38

Limits and Infinity


Find the limits in Exercises 37–46.


2x² + 3
lim -------------
x→⁻∞ 5x² + 7

Verified step by step guidance
1
Identify the highest degree terms in the numerator and the denominator. In this case, both the numerator and the denominator have the highest degree term of x².
Rewrite the limit expression by factoring out the highest degree term from both the numerator and the denominator. This gives us: lim (x→⁻∞) [(2x²/x²) + (3/x²)] / [(5x²/x²) + (7/x²)].
Simplify the expression by canceling out the x² terms in the numerator and the denominator. This results in: lim (x→⁻∞) [2 + (3/x²)] / [5 + (7/x²)].
Evaluate the limit as x approaches negative infinity. As x becomes very large in magnitude, the terms (3/x²) and (7/x²) approach zero.
Conclude that the limit is the ratio of the leading coefficients of the highest degree terms, which is 2/5.

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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. They help in understanding the behavior of functions at specific points, including points of discontinuity or infinity. In this context, we are interested in the limit of a rational function as x approaches negative infinity.
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One-Sided Limits

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials. They can exhibit different behaviors as the variable approaches certain values, including infinity. Analyzing the leading terms of the numerator and denominator is crucial for determining the limit of a rational function at infinity.
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Intro to Rational Functions

Behavior at Infinity

The behavior of functions as they approach infinity is essential for evaluating limits. For rational functions, this often involves comparing the degrees of the polynomials in the numerator and denominator. If the degrees are the same, the limit is the ratio of the leading coefficients; if the degree of the numerator is less, the limit approaches zero.
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Cases Where Limits Do Not Exist
Related Practice
Textbook Question

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limt→0 sin(1 − cos t) / (1 − cos t)

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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

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


y = √(2x + 3)

Textbook Question

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limh→0 sin(sin h) / sin h

Textbook Question

Limits as x → ∞ or x → −∞


The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x. Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23–36. Write ∞ or −∞ where appropriate.


lim x→∞ (2√x + x⁻¹) / (3x − 7)