Textbook Question
Domains and Asymptotes
Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.
y = 4 + 3x² / (x² + 1)
336
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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 2.25
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
Verified step by step guidance1
First, understand the problem: We need to find the limit of the function (4g(x))^(1/3) as x approaches 0, and we know that this limit equals 2.
To solve this, we can start by setting up the equation based on the given limit: lim (4g(x))^(1/3) = 2 as x → 0.
Next, cube both sides of the equation to eliminate the cube root. This gives us: lim 4g(x) = 2^3 = 8 as x → 0.
Now, solve for g(x) by dividing both sides of the equation by 4: lim g(x) = 8/4 = 2 as x → 0.
Finally, verify that the limit of g(x) as x approaches 0 is indeed 2, which satisfies the original condition of the problem.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits
A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It helps in understanding how functions behave near specific points, which is crucial for defining continuity and derivatives. In this case, we are interested in the limit of the function g(x) as x approaches 0.
Recommended video:
05:50
One-Sided Limits
Continuous Functions
A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. This concept is essential when evaluating limits, as it allows us to substitute values directly into the function if it is continuous. If g(x) is continuous at x = 0, we can directly evaluate the limit without any complications.
Recommended video:
05:34Intro to Continuity
Cube Root Function
The cube root function, denoted as (4g(x))^(1/3), is a transformation of the function g(x) that involves taking the cube root of the product of 4 and g(x). Understanding how this function behaves as x approaches 0 is key to solving the limit problem. The limit provided indicates that as x approaches 0, the cube root of 4g(x) approaches 2, which can help us find the value of g(0).
Recommended video:
5:57Graphs of Common Functions
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)
310
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Textbook Question
At what points are the functions in Exercises 13–30 continuous?
y = √(2x + 3)
226
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Textbook Question
Limits and Infinity
Find the limits in Exercises 37–46.
x⁴ + x³
lim -----------------
x→∞ 12x³ + 128
303
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Textbook Question
Limits of Average Rates of Change
Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form limh→0 (f(x+h) − f(x)) / h occur frequently in calculus. In Exercises 57–62, evaluate this limit for the given value of x and function f.
f(x) = x², x = 1
333
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Textbook Question
Limits with trigonometric functions
Find the limits in Exercises 43–50.
limx→−π √(x + 4) cos(x + π)
262
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