Question

Let \( f: \mathbb{R} \to \mathbb{R} \) \(\text{ be any function defined as }\) \[ f(x) = \begin{cases} x^\alpha \sin \left( \frac{1}{x^\beta} \right) & \text{for } x \neq 0, \\ 0 & \text{for } x = 0, \end{cases} \] where \( \alpha, \beta \in \mathbb{R} \). Which of the following is true? \( \mathbb{R} \) denotes the set of all real numbers.

• A. \( f(x) \) is continuous at \( x = 0 \), for all \( \alpha \) and \( \beta \in \mathbb{R} \)
• B. \( f(x) \) is differentiable at \( x = 0 \) for all \( \alpha > 0 \) and \( \beta > 0 \)
• C. \( f(x) \) is continuous at \( x = 10 \) for only \( \alpha > 0 \) and \( \beta > 0 \)
• D. \( f(x) \) is continuous at \( x = 0 \) for all \( \alpha > 0 \) and \( \beta \in \mathbb{R} \)

Hint:

For continuity at \( x = 0 \), the function must approach the value at 0 as \( x \to 0 \). This is determined by the behavior of the leading term as \( x \to 0 \).

Answer 1

The Correct Option is D

We are given the function \( f(x) \), and we need to determine the conditions under which it is continuous at \( x = 0 \).

Step 1: Investigating continuity at \( x = 0 \).
For the function to be continuous at \( x = 0 \), the limit of \( f(x) \) as \( x \to 0 \) must exist and equal the value of the function at \( x = 0 \), which is 0.

Step 2: Compute the limit.
We analyze the behavior of \( f(x) = x^\alpha \sin \left( \frac{1}{x^\beta} \right) \) as \( x \to 0 \). The sine term oscillates between \( -1 \) and \( 1 \), so the limit of \( f(x) \) is controlled by the term \( x^\alpha \). For \( f(x) \) to approach 0 as \( x \to 0 \), we need \( \alpha > 0 \). Thus, \( f(x) \) is continuous at \( x = 0 \) for all \( \alpha > 0 \) and \( \beta \in \mathbb{R} \). Therefore, the correct answer is option (d).