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
Continuity
Problem 32
Textbook Question
Continuous Extension
Explain why the function ƒ(𝓍) = sin(1/𝓍) has no continuous extension to 𝓍 = 0.

1
To understand why the function ƒ(𝓍) = sin(1/𝓍) has no continuous extension to 𝓍 = 0, we first need to consider the behavior of the function as 𝓍 approaches 0. The function is defined for all 𝓍 ≠ 0, but we are interested in its behavior as 𝓍 gets very close to 0.
As 𝓍 approaches 0, the expression 1/𝓍 becomes very large in magnitude, which means that the argument of the sine function oscillates rapidly between positive and negative values. This rapid oscillation causes the function ƒ(𝓍) = sin(1/𝓍) to oscillate between -1 and 1 without settling down to any particular value.
For a function to have a continuous extension at a point, the limit of the function as it approaches that point must exist and be finite. In this case, we need to check if the limit of ƒ(𝓍) as 𝓍 approaches 0 exists.
To determine the limit, consider the fact that for any sequence of values of 𝓍 approaching 0, the corresponding sequence of values of 1/𝓍 will cover all real numbers densely. This means that the values of sin(1/𝓍) will cover the interval [-1, 1] densely as well, without converging to a single value.
Since the limit of ƒ(𝓍) as 𝓍 approaches 0 does not exist (the function does not approach a single value), there is no way to define ƒ(0) such that the function becomes continuous at 𝓍 = 0. Therefore, ƒ(𝓍) = sin(1/𝓍) has no continuous extension to 𝓍 = 0.

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
6mPlay a video:
Was this helpful?
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. For the function ƒ(𝓍) = sin(1/𝓍), we need to analyze the limit as 𝓍 approaches 0. If the limit does not exist or is not finite, the function cannot be continuously extended to that point.
Recommended video:
One-Sided Limits
Continuity
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. For ƒ(𝓍) = sin(1/𝓍}, we find that as 𝓍 approaches 0, the function oscillates between -1 and 1, indicating that it does not settle at a single value, thus failing the continuity requirement at 𝓍 = 0.
Recommended video:
Intro to Continuity
Oscillation
Oscillation refers to the behavior of a function that fluctuates between values without converging to a single limit. In the case of ƒ(𝓍) = sin(1/𝓍), as 𝓍 approaches 0, the function oscillates infinitely between -1 and 1, which means it does not approach any specific value, preventing a continuous extension to 𝓍 = 0.
Recommended video:
Cases Where Limits Do Not Exist
Related Videos
Related Practice