Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
1. Limits and Continuity
Finding Limits Algebraically
Problem 2.25b
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

1
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.

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
2mPlay a video:
Was this helpful?
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:
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:
Intro 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:
Graphs of Common Functions
Watch next
Master Finding Limits by Direct Substitution with a bite sized video explanation from Callie
Start learningRelated Videos
Related Practice