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Part CCSIR NET December 2023types-of-discontinuity

Types of discontinuity

Which of the following statements are true?

  1. A.f(x) = [x] sin(1/x) for x ≠ 0, f(0) = 0 (where [x] is the integer part) has a discontinuity at 0 which is removable.
  2. B.f:[0,∞)→R,f(x)=sin⁡(log⁡x)f : [0,\infty) \to \mathbb{R}, f(x) = \sin(\log x) for x ≠ 0, f(0) = 0, has a discontinuity at 0 which is NOT removable.✓ Correct
  3. C.f(x) = e1/xe^{1/x} for x < 0, f(x) = e1/(x+1)e^{1/(x+1)} for x ≥ 0, has a jump discontinuity at 0.✓ Correct
  4. D.If f,g:[0,1]→Rf, g : [0,1] \to \mathbb{R} are of bounded variation, then fg has at most countably many discontinuities.✓ Correct

The idea, in plain words

A break in a graph can be repaired only if the function heads towards a single value from both sides. Check each one from the left and from the right separately: some settle on two different values, and some never settle at all because they oscillate faster and faster.

Solution

(1) As x → 0⁻, [x] = −1 and −sin(1/x) oscillates: no left limit, not removable.

(2) sin(log x) oscillates as x → 0⁺.

(3) Left limit 0, right limit e.

(4) A product of BV functions is BV, and BV functions have only countably many (jump) discontinuities.

Why each option is right or wrong

Checked against 2 symbolic computations
  • A.WrongBoundary and endpoint

    False.

    Check both sides

    because [x] jumps at 0.

    For 0 < x < 1 the integer part is 0

    so f ≡ 0 on the right and the right limit is 0 — which is where most students stop.

    For −1 ≤ x < 0 the integer part is −1

    so f(x) = −sin(1/x)

    which oscillates without limit as x → 0⁻.

    A removable discontinuity needs the two-sided limit to exist; here the left limit does not.

    The mistake: evaluating the side on which the floor function is harmless and never checking the other.

  • B.Correct

    True.

    As x → 0⁺, log⁡x→−∞\log x \to -\infty

    so sin(log x) oscillates through [−1, 1] forever and no limit exists.

    A discontinuity is removable only when the limit exists but disagrees with the assigned value; here there is nothing to remove.

  • C.Correct

    True, and it is a genuine jump because both one-sided limits exist and differ.

    From the left, 1/x→−∞1/x \to -\infty so e^(1/x) → 0 (SymPy samples e^(1/t) down to t = −0.05 and reaches 2×10⁻9)^{9}).

    From the right, e(1/(x+1))→e1=e≈2.718e^(1/(x+1)) \to e^{1} = e \approx 2.718.

    Left limit 0, right limit e.

  • D.Correct

    True.

    A product of two functions of bounded variation is again of bounded variation on [0,1] (each is a difference of monotone functions, and BV is closed under products of bounded functions).

    A BV function has only jump discontinuities and at most countably many of them.

1 of these 1 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: Continuous on a bounded interval ⇒ bounded

More on this topic

The chapter behind this: Continuity vs uniform continuity vs Lipschitz — free to read

From Continuity and Differentiation › Continuity, uniform continuity, Lipschitz

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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