Question

Determine the largest interval over which the given function is continuous. $$ F(x)=x \sin \frac{1}{x} ; c=0 $$

Short answer

The largest interval where \( F(x) \) is continuous is \( \mathbb{R} \).

Step-by-step solution

Understanding Continuity

A function is continuous on an interval if it is continuous at every point in that interval. To determine where the given function is continuous, we first need to understand the function: \( F(x) = x \sin \frac{1}{x} \). This function has an issue at \( x = 0 \) as \( \frac{1}{x} \) becomes undefined there.

Analyzing Function Behavior at \(x \neq 0\)

For \( x eq 0 \), the term \( \sin \frac{1}{x} \) is well-defined since \( \frac{1}{x} \) simply results in a real number. Moreover, \( x \sin \frac{1}{x} \) is a product of continuous functions (polynomial \( x \) and \( \sin \)), ensuring the function is continuous for all \( x eq 0 \).

Evaluating the Continuity at \(x = 0\)

Evaluate the continuity at \( x = 0 \) by checking the limit of \( F(x) \) as \( x \) approaches 0. The limit \( \lim_{x \to 0} x \sin \frac{1}{x} = 0 \) since \( \sin \) is bounded between -1 and 1, making \( x \sin \frac{1}{x} \) approach 0 as \( x \to 0 \).

Concluding Continuity on \( \mathbb{R} \)

Since the limit at \( x = 0 \) equals \( F(0) \) (assuming \( F(0) = 0 \)), the function is continuous at \( x = 0 \). Combining this with continuity for \( x eq 0 \), we conclude that \( F(x) \) is continuous over its entire domain, which is all real numbers \( \mathbb{R} \).

Key concepts

Interval of Continuity

When we talk about the interval of continuity, we refer to the range of input values for which a function is smooth, without any breaks, jumps, or undefined points.
For the function \( F(x) = x \sin \frac{1}{x} \), we must look carefully at its behavior across its domain.
- **Understanding the Domains:** Generally, most functions are defined for all real numbers unless they include a division by zero or other undefined operations. Here, \( \frac{1}{x} \) could be problematic at \( x = 0 \).- **Identifying Issues:** For our function, \( \sin \frac{1}{x} \) becomes undefined specifically at \( x = 0 \) due to the division.- **Solution:** However, away from \( x = 0 \), there are no disruptions, meaning \( x \sin \frac{1}{x} \) is continuous in all other places.By considering these factors, we find that the function is continuous on the entire real line \( \mathbb{R} \), if we can prove continuity at \( x = 0 \) itself.

Limit and Continuity

To establish a function's continuity, especially at a point where the function's definition is tricky, limits are invaluable.
In our exercise, we determine the continuity at \( x = 0 \) using limits.
- **Limit Approach:** We check \( \lim_{x \to 0} F(x) = \lim_{x \to 0} x \sin \left( \frac{1}{x} \right) \).- **Conclusion of the Limit:** Since \( \sin \) is bounded between -1 and 1, multiplying by \( x \) (which approaches 0) drives the expression to 0.This calculation shows that not only does the limit exist at 0, but it equals 0. Consequently, since \( F(0) = 0 \) as well, \( F(x) \) remains continuous at this tricky point.

Function Behavior at a Point

The behavior of a function at a specific point, like where it might seemingly become undefined, tells us a lot about its continuity.
For \( x = 0 \), \( F(x) = x \sin \frac{1}{x} \) might look misleading since \( \sin \frac{1}{x} \) cycles rapidly as \( x \) nears 0.
- **Analytical Observation:** Though \( \sin \) wildly oscillates, each oscillation remains bounded, and \( x \, \sin \) limits this oscillation to disappear.- **Visualizing the Behavior:** As \( x \) nears 0, the function behaves more predictably closer to a calm alignment.- **Evaluated Value:** Putting \( F(0) = 0 \) confirms the function's calmness there.Hence, the function behaves nicely at 0, transforming seeming chaos into order, ensuring it fits smoothly without a hitch.