Physics I / Limits and continuity
Practice question · Sort into groups

Sort each function by what happens at the marked point.

Groups: Continuous there · Removable hole · Jump · Vertical asymptote

Hints
  1. Try to cancel first: (x24)/(x2)(x^{2}-4)/(x-2) simplifies everywhere except one missing point.
  2. x|x| has a sharp corner at 0, but sharp is not broken. Trace it without lifting your pen.
Show the answer

Continuous there: x|x| at x=0x = 0

Removable hole: x24x2\frac{x^{2}-4}{x-2} at x=2x = 2

Jump: Postage cost as weight crosses a price step

Vertical asymptote: 1/x1/x at x=0x = 0

Why

The three ways continuity fails: a hole (limit exists, point missing), a jump (one-sided limits disagree), a blow-up (no finite limit). The corner of x|x| breaks none of these, corners are a problem for derivatives, not for continuity.

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