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
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
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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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