Extension to closed interval

• A.WrongStandard counterexample

  False. sin(1/x) has no limit as x → 0⁺ — SymPy samples it on (0.001, 0.05] and finds values ranging down to −0.99998 and up near +1

  oscillating without settling.

  A continuous function on a bounded interval is uniformly continuous exactly when it extends continuously to the closure, and this one cannot.

• B.Correct

  True.

  SymPy confirms $e^{-1/x^{2}}$ → 0 as x → 0⁺, and the function is already continuous at x = 1

  so it extends continuously to [0,1].

  A continuous function on a compact interval is uniformly continuous by Heine–Cantor, and restricting back to (0,1) preserves that.

  The trigger: on a bounded open interval, uniform continuity is exactly continuous extendability to the endpoints — check the one-sided limits and nothing else.

• C.WrongStandard counterexample

  False.

  The factor eˣ tends to 1 at x = 0 and is harmless

  but cos(1/x) oscillates exactly as sin(1/x) does.

  Multiplying an oscillating factor by a well-behaved one leaves the oscillation intact.

• D.WrongStandard counterexample

  False, for the same reason. cos x → 1 as x → 0

  while $\cos(\pi/x)$ oscillates without limit — the $\pi$ in the numerator changes the frequency but not the behaviour.

  Three of the four options oscillate at 0; the exponential in (B) is the only one that damps.

3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.