Step 1: Understanding the Concept:
We need to analyze the properties of the function \(f(x) = \sin(\frac{1}{x})\) on the specified interval \((0, 2\pi]\). The two properties to check are boundedness and continuity.
Step 2: Key Formula or Approach:
- Boundedness: A function \(f\) is bounded on an interval \(I\) if there exists a real number \(M > 0\) such that \(|f(x)| \le M\) for all \(x \in I\).
- Continuity: A function is continuous on an interval if it is continuous at every point in that interval. The function \(\sin(u)\) is continuous for all real \(u\), and the function \(g(x) = 1/x\) is continuous for all \(x \neq 0\). The composition of continuous functions is continuous.
Step 3: Detailed Explanation:
1. Boundedness:
The function is \(f(x) = \sin(\frac{1}{x})\). The range of the sine function, \(\sin(u)\), is always \([-1, 1]\), regardless of its argument \(u\).
Therefore, for any \(x\) in the interval \((0, 2\pi]\), the value of \(\frac{1}{x}\) will be some real number, and the sine of that number will be between -1 and 1.
\[ -1 \le \sin\left(\frac{1}{x}\right) \le 1 \]
This means \(|f(x)| \le 1\) for all \(x\) in the given interval.
So, the function is bounded.
2. Continuity:
The function \(f(x)\) is a composition of two functions: \(g(x) = \frac{1}{x}\) and \(h(u) = \sin(u)\), such that \(f(x) = h(g(x))\).
- The function \(g(x) = \frac{1}{x}\) is continuous for all \(x \neq 0\). The given interval is \((0, 2\pi]\), which does not include 0. So, \(g(x)\) is continuous on this interval.
- The function \(h(u) = \sin(u)\) is continuous for all real numbers \(u\).
- The composition of two continuous functions is continuous. Since \(g(x)\) is continuous on \((0, 2\pi]\) and \(h(u)\) is continuous everywhere, their composition \(f(x) = \sin(\frac{1}{x})\) is continuous on the interval \((0, 2\pi]\).
Conclusion:
The function \(f(x) = \sin(\frac{1}{x})\) is both bounded and continuous on the interval \((0, 2\pi]\).
Step 4: Final Answer:
The function is bounded and continuous, which corresponds to option (D).