Answer: f(x) = (x 2 -1)/(x-1) at x = 1, undefined there but limit exists.
- A f(x) = 1/x at x = 0, an infinite discontinuity that grows without bound
- B f(x) = [x] at integer points, a jump discontinuity in the step function
- C f(x) = (x<sup>2</sup>-1)/(x-1) at x = 1, undefined there but limit exists
- D f(x) = x<sup>2</sup> everywhere, a polynomial that is already smooth and continuous
Correct answer: C. f(x) = (x<sup>2</sup>-1)/(x-1) at x = 1, undefined there but limit exists
Explanation: (x<sup>2</sup>-1)/(x-1) simplifies to x+1 for x not equal to 1, so the limit as x approaches 1 exists (equals 2) even though f(1) is undefined; redefining f(1)=2 removes the gap.
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.