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 38
Textbook Question
Evaluate each limit and justify your answer.
lim x→0 (x / √16x+1-1)^1/3

1
Step 1: Recognize that the limit involves a form that may be indeterminate as x approaches 0. The expression inside the limit is (x / (\sqrt{16x+1} - 1))^{1/3}.
Step 2: Simplify the expression inside the limit. Notice that the denominator \sqrt{16x+1} - 1 can be rationalized. Multiply the numerator and the denominator by the conjugate \sqrt{16x+1} + 1.
Step 3: After multiplying by the conjugate, the denominator becomes a difference of squares: (\sqrt{16x+1})^2 - 1^2 = 16x. The expression now is x(\sqrt{16x+1} + 1) / 16x.
Step 4: Simplify the expression further by canceling x in the numerator and denominator, resulting in (\sqrt{16x+1} + 1) / 16.
Step 5: Evaluate the limit as x approaches 0. Substitute x = 0 into the simplified expression to find the limit of the original expression.

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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 points of interest, including points of discontinuity or indeterminate forms. Evaluating limits often involves techniques such as substitution, factoring, or applying L'Hôpital's rule when faced with indeterminate forms.
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Indeterminate Forms
Indeterminate forms arise in calculus when evaluating limits that do not lead to a clear numerical result, such as 0/0 or ∞/∞. These forms require further analysis or manipulation to resolve. Recognizing an indeterminate form is crucial for applying appropriate techniques, such as algebraic simplification or L'Hôpital's rule, to find the limit's value.
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Cube Roots
The cube root function, denoted as (x)^(1/3), is the inverse operation of cubing a number. It is important in limit problems involving expressions raised to fractional powers. Understanding how cube roots behave, especially near critical points like zero, is essential for accurately evaluating limits that involve such expressions, as they can affect the continuity and differentiability of the function.
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