Question

Let \[ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \] Then which of the following is true:

• A. \( f(x) \) is not continuous at \( x = 0 \)
• B. \( f(x) \) is not differentiable at \( x = 0 \)
• C. \( f'(x) \) is not continuous at \( x = 0 \)
• D. \( f'(x) \) is continuous at \( x = 0 \)

Hint:

Remember to carefully apply the definitions of continuity and differentiability at a point, especially for piecewise functions.

Answer 1

The Correct Option is D

Step 1: Check for continuity of \( f(x) \) at \( x = 0 \).
For \( f(x) \) to be continuous at \( x = 0 \), we need \( \lim_{x \to 0} f(x) = f(0) \).
We know that \( f(0) = 0 \). Let's find the limit: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\frac{1}{x} \] Since \( -1 \leq \sin\frac{1}{x} \leq 1 \), we have \( -x^2 \leq x^2 \sin\frac{1}{x} \leq x^2 \).
By the Squeeze Theorem, \( \lim_{x \to 0} x^2 \sin\frac{1}{x} = 0 \). Since \( \lim_{x \to 0} f(x) = 0 = f(0) \), \( f(x) \) is continuous at \( x = 0 \). 
Step 2: Check for differentiability of \( f(x) \) at \( x = 0 \).
The derivative at \( x = 0 \) is given by: \[ f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin\frac{1}{h} - 0}{h} = \lim_{h \to 0} h \sin\frac{1}{h} \] Since \( -1 \leq \sin\frac{1}{h} \leq 1 \), we have \( -|h| \leq h \sin\frac{1}{h} \leq |h| \).
By the Squeeze Theorem, \( \lim_{h \to 0} h \sin\frac{1}{h} = 0 \). So, \( f'(0) = 0 \), which means \( f(x) \) is differentiable at \( x = 0 \). 
Step 3: Find the derivative \( f'(x) \) for \( x \neq 0 \).
Using the product rule and the chain rule: \[ f'(x) = \frac{d}{dx} \left( x^2 \sin\frac{1}{x} \right) = 2x \sin\frac{1}{x} + x^2 \left( \cos\frac{1}{x} \cdot (-\frac{1}{x^2}) \right) = 2x \sin\frac{1}{x} - \cos\frac{1}{x} \] Step 4: Check for continuity of \( f'(x) \) at \( x = 0 \).
For \( f'(x) \) to be continuous at \( x = 0 \), we need \( \lim_{x \to 0} f'(x) = f'(0) \). \[ \lim_{x \to 0} f'(x) = \lim_{x \to 0} \left( 2x \sin\frac{1}{x} - \cos\frac{1}{x} \right) \] We know that \( \lim_{x \to 0} 2x \sin\frac{1}{x} = 0 \). However, \( \lim_{x \to 0} \cos\frac{1}{x} \) does not exist.
Therefore, \( \lim_{x \to 0} f'(x) \) does not exist, which means \( f'(x) \) is not continuous at \( x = 0 \).