Define f(x) = (x²-4)/(x-2) for x ≠ 2, and f(2) = 5. What type of discontinuity does f have at x = 2?
AJump discontinuity
BInfinite discontinuity
CRemovable discontinuity
DNone — f is continuous at x = 2
Factor: (x²-4)/(x-2) = (x+2)(x-2)/(x-2) = x+2 for x ≠ 2. So lim(x→2) f(x) = 4, which exists. But f(2) = 5 ≠ 4 — condition 3 fails. This is a removable discontinuity (a displaced point). If f(2) were redefined as 4, the discontinuity would be removed.
Question 2 True / False
A function that is defined at nearly every real number is expected to be continuous at nearly every real number.
TTrue
FFalse
Answer: False
Being defined everywhere satisfies only condition 1. The limit must also exist (condition 2) and equal the function value (condition 3). For example, f(x) = 0 for x ≠ 0 and f(0) = 1 is defined everywhere, but at x = 0 the limit is 0 while f(0) = 1 — condition 3 fails, so f is discontinuous at 0.
Question 3 Short Answer
State the three conditions for continuity at x = a, and describe geometrically what each one rules out.
Think about your answer, then reveal below.
Model answer: (1) f(a) is defined — rules out a hole (missing point). (2) lim(x→a) f(x) exists — rules out a jump (left and right limits disagree). (3) The limit equals f(a) — rules out a displaced point (limit exists but the function value is shifted away from it).
Each condition targets exactly one type of discontinuity. Together they guarantee the graph passes through (a, f(a)) smoothly with no break of any kind. Violating exactly one condition identifies the discontinuity type precisely.