Answer: Yes, since x sin(1/x) is squeezed between -|x| and |x| which go to 0.
- A No, the limit does not exist since sin(1/x) oscillates without settling
- B Yes, since x sin(1/x) is squeezed between -|x| and |x| which go to 0
- C No, because sin(1/x) oscillates infinitely often near x = 0
- D Cannot be determined without evaluating the one-sided limits separately
Correct answer: B. Yes, since x sin(1/x) is squeezed between -|x| and |x| which go to 0
Explanation: Since |x sin(1/x)| <= |x|, by the squeeze theorem the limit as x approaches 0 is 0, which equals f(0), so f is continuous at 0.
f(x)=|x| is perfectly continuous at x=0 (no break, no jump), but the curve has a sharp corner there: approaching from the left gives slope -1 and from the right gives slope +1 - since these one-sided derivatives disagree, the function is not differentiable at that point.
Concept context
When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.