Question

The function $f(x) = \begin{cases} x \sin\frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is}

• A. continuous everywhere
• B. continuous only at $x = 0$
• C. not continuous at $x = 0$
• D. nowhere continuous

Hint:

Functions of the form $x^p \sin(1/x)$ at $x=0$ (with $f(0)=0$) are:
- Continuous if $p > 0$.
- Differentiable if $p > 1$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question tests the continuity of a piecewise-defined real function, specifically focusing on its behavior near the boundary point $x = 0$.

Step 2: Key Formula or Approach: A function $f(x)$ is continuous at $x = c$ if:
\[ \lim_{x \to c} f(x) = f(c) \] For $x \neq 0$, the function is continuous because it is the product of two continuous functions ($x$ and $\sin(1/x)$).
At $x = 0$, we must evaluate the limit using the Squeeze Theorem.

Step 3: Detailed Explanation:

• For $x \neq 0$, the function is defined as $f(x) = x \sin(1/x)$.

• Since the linear function $y = x$ and the trigonometric function $y = \sin(1/x)$ are both continuous on $(-\infty, 0) \cup (0, \infty)$, their product $f(x)$ is continuous everywhere except possibly at $x = 0$.

• Let's test the continuity at $x = 0$ by evaluating the limit:
\[ \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \]
• We know that the sine function is always bounded between $-1$ and $1$ for all real arguments:
\[ -1 \le \sin\left(\frac{1}{x}\right) \le 1 \]
• Multiplying this inequality by $|x|$:
\[ -|x| \le x \sin\left(\frac{1}{x}\right) \le |x| \]
• Let's evaluate the limit of the outer terms as $x \to 0$:
\[ \lim_{x \to 0} -|x| = 0 \quad \text{and} \quad \lim_{x \to 0} |x| = 0 \]
• By the Squeeze Theorem, the limit of the middle term must also be 0:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0 \]
• Since the limit of $f(x)$ as $x \to 0$ matches the defined function value $f(0) = 0$, the function is continuous at $x = 0$.

• Therefore, the function is continuous for all real numbers.

Step 4: Final Answer:
The function is continuous everywhere, which is option (A).