Question

If \[ f(x) = \begin{cases} x^\alpha \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases} \] Which of the following is true?

• A. \( f(x) \) is continuous and differentiable if \( 0 \leq \alpha<1 \)
• B. \( f(x) \) is discontinuous and not differentiable if \( 0 \leq \alpha<1 \)
• C. \( f(x) \) is continuous and differentiable for \( \alpha<1 \)
• D. \( f(x) \) is discontinuous and differentiable for \( \alpha<1 \)

Hint:

For continuity, check \( \lim_{x \to a} f(x) = f(a) \). For differentiability, compute \( \lim_{x \to a} \frac{f(x) - f(a)}{x-a} \).

Answer 1

The Correct Option is C

Step 1: Checking Continuity at \( x = 0 \) \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^\alpha \sin \left(\frac{1}{x}\right) \] Since \( -1 \leq \sin(1/x) \leq 1 \), multiplying by \( x^\alpha \): \[ - x^\alpha \leq x^\alpha \sin(1/x) \leq x^\alpha \] Taking limits, \( \lim_{x \to 0} f(x) = 0 \), which equals \( f(0) \). So, \( f(x) \) is continuous. 

Step 2: Checking Differentiability at \( x = 0 \) Differentiating, \[ f'(x) = \alpha x^{\alpha-1} \sin(1/x) - x^{\alpha-2} \cos(1/x) \] For \( f'(0) \) to exist, \( \alpha<1 \) is needed to make the second term vanish.