Correct Answer: D ($f(x)$ is continuous at $x=0$ for all $\alpha > 0$ & $\beta \in \mathbb{R}$)

To determine which statement is true, we need to find the conditions on $\alpha$ and $\beta$ that ensure the function $f(x)$ is continuous at $x=0$.

Step 1: The Definition of Continuity at a Point

For a function $f(x)$ to be continuous at $x=0$, the limit of the function as $x$ approaches 0 must exist and be equal to the function's value at 0.

$$ \lim_{x \to 0} f(x) = f(0) $$

From the problem definition, we are given that $f(0) = 0$. So, we need to find the conditions under which:

$$ \lim_{x \to 0} x^\alpha \sin\left(\frac{1}{x^\beta}\right) = 0 $$

Step 2: Applying the Squeeze Theorem

This type of limit is a classic application of the Squeeze Theorem. We know that the sine function is always bounded between -1 and 1, regardless of its argument. Therefore:

$$ -1 \le \sin\left(\frac{1}{x^\beta}\right) \le 1 $$

This holds true for any real value of $\beta$. Now, we multiply the inequality by $|x^\alpha|$:

$$ -|x^\alpha| \le \left|x^\alpha \sin\left(\frac{1}{x^\beta}\right)\right| \le |x^\alpha| $$

The Squeeze Theorem states that if the outer two functions approach the same limit, the function in the middle must also approach that limit. For our limit to be 0, we need:

$$ \lim_{x \to 0} |x^\alpha| = 0 $$

Step 3: Determining the Conditions on $\alpha$ and $\beta$

The limit $\lim_{x \to 0} |x^\alpha|$ will be equal to 0 if and only if the exponent $\alpha$ is strictly greater than 0.

• If $\alpha > 0$, then as $x$ gets closer to 0, $x^\alpha$ also gets closer to 0.
• If $\alpha = 0$, the limit is 1.
• If $\alpha < 0$, the limit is $\infty$.

Therefore, for the function to be continuous at $x=0$, we must have $\alpha > 0$.

Our analysis did not place any restrictions on $\beta$, because the sine term is always bounded. Thus, $\beta$ can be any real number ($\beta \in \mathbb{R}$).

Conclusion for Continuity: $f(x)$ is continuous at $x=0$ for all $\alpha > 0$ and for all $\beta \in \mathbb{R}$.

Step 4: Evaluating the Options

• A. $f(x)$ is differentiable for all $\alpha>0\&\beta>0$. (False - Differentiability requires $\alpha > 1$).
• B. $f(x)$ is continuous for only $\alpha>0\&\beta>0$. (False - It is continuous for all real $\beta$).
• C. $f(x)$ is continuous for all $\alpha\&\beta \in \mathbb{R}$. (False - It requires $\alpha>0$).
• D. $f(x)$ is continuous at $x=0$ for all $\alpha>0\&\beta \in \mathbb{R}$. (True - This matches our derived condition).