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MHT-CET Maths · Limits

Discontinuities of [x], |x| and sgn x — Counting the Points

The greatest-integer, modulus and sign functions carry built-in jumps — the question is where they land, whether another factor cancels them, and how many there are in a given interval.

Why this matters

The smallest page in the chapter — 10 PYQs at 40% HARD — but a distinct question type with no coverage anywhere else: 'how many points of discontinuity' and 'discontinuous at which set'. It reuses the one-sided habit from the first page and the three-part test from the continuity pages; what is new is the counting, and the one case where a vanishing factor swallows a jump. One of them carries an exam key that contradicts the mathematics, and knowing which is part of the preparation.

Concept 1 of 5: [x] Is Discontinuous at Every Integer — Counting Them

The staircase [x][x] steps up by 11 at each integer and is flat in between. So its discontinuities are exactly the integers, and 'how many in (a,b)(a, b)' is a counting question: how many integers lie strictly inside.

Definition

  • At an integer nn: left limit n−1n - 1, right limit nn, value nn. The left limit disagrees, so [x][x] is discontinuous at every integer and continuous everywhere else.
  • [x][x] is right-continuous at integers (right limit == value). At a closed left endpoint that is an integer, there is no left approach, so no discontinuity there; at a closed right endpoint that is an integer, the left approach fails, so it counts.
  • Counting integers in an open interval (a,b)(a, b): from ⌈a⌉\lceil a \rceil (or a+1a + 1 if aa is an integer) to ⌊b⌋\lfloor b \rfloor (or b−1b - 1 if bb is an integer). In (−72,100)\left(-\frac{7}{2}, 100\right): −3,−2,…,99-3, -2, \dots, 99 — that is 3+1+99=1033 + 1 + 99 = 103.
  • [kx][kx] jumps where kxkx is an integer, i.e. at multiples of 1k\dfrac{1}{k}; [x2]\left[\dfrac{x}{2}\right] jumps at even integers.

Jumps of the greatest-integer function

lim⁡x→n−[x]=n−1≠n=[n](n∈Z)[kx] jumps at x=mk, m∈Z\lim_{x\to n^-}[x] = n - 1 \ne n = [n] \quad (n \in \mathbb{Z}) \qquad [kx] \text{ jumps at } x = \frac{m}{k},\ m \in \mathbb{Z}
xy[x] = n on [n, n+1) · jumps up by 1 at each integer

Worked example

How many points of discontinuity does f(x)=[x]f(x) = [x] have on (−2.5,6)(-2.5, 6)?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q142Moderate

Example 1 · Limits · Discontinuities of [x], |x| and sgn x — Counting the Points

The number of discontinuities of the greatest integer function f(x)=[x]f(x)=[x], x∈(−72,100)x\in\left(-\frac{7}{2},100\right)

Miscounting the negative side

In (−72,100)\left(-\frac{7}{2}, 100\right) the integers are −3,−2,−1-3, -2, -1 (three of them), then 00, then 11 to 9999. Forgetting 00, or including 100100, gives 102102 or 104104 — both offered.

Concept 2 of 5: Signum-Type Jumps: (x − a)/|x − a| and Products with It

x−a∣x−a∣\dfrac{x - a}{|x - a|} is +1+1 on one side of aa and −1-1 on the other: a jump of size 22 that no choice of f(a)f(a) can bridge. Anything multiplied by it inherits the jump unless the other factor is 00 at aa.

Definition

  • x−a∣x−a∣=∣x−a∣x−a={1,x>a−1,x<a\dfrac{x - a}{|x - a|} = \dfrac{|x - a|}{x - a} = \begin{cases} 1, & x > a \\ -1, & x < a \end{cases}; it is undefined at aa and discontinuous there whatever value is assigned.
  • p(x)∣(x−1)(x−2)∣\dfrac{p(x)}{|(x - 1)(x - 2)|} with pp divisible by (x−1)(x−2)(x - 1)(x - 2): the polynomial parts cancel, leaving sign factors at 11 and 22. Each gives a jump, so ff is continuous on R−{1,2}\mathbb{R} - \{1, 2\} and nowhere else can be repaired.
  • A piecewise function built from x−3∣x−3∣+a\dfrac{x - 3}{|x - 3|} + a on the left and ∣x−3∣x−3+b\dfrac{|x - 3|}{x - 3} + b on the right can be continuous — the sign factors evaluate to −1-1 and +1+1, giving ordinary equations −1+a=1+b=f(3)-1 + a = 1 + b = f(3).
  • ∣x∣|x| itself is continuous everywhere; only ∣x∣|x| divided by something vanishing at the same point produces a jump.

The sign factor

x−a∣x−a∣={1,x>a−1,x<ajump of size 2 at a\frac{x - a}{|x - a|} = \begin{cases} 1, & x > a \\ -1, & x < a \end{cases} \qquad \text{jump of size } 2 \text{ at } a

Worked example

f(x)=x2−1∣x−1∣f(x) = \dfrac{x^2 - 1}{|x - 1|} for x≠1x \neq 1. Can f(1)f(1) be chosen so that ff is continuous at 11?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q102Moderate

Example 2 · Limits · Discontinuities of [x], |x| and sgn x — Counting the Points

Let f(x)={x4−5x2+4∣(x−1)(x−2)∣,x≠1,26,x=112,x=2f(x) =\left\{ \begin{matrix} \frac{x^{4}- 5x^{2}+ 4}{|(x- 1)(x- 2)|} & ,x\neq 1,2 \\ 6 & ,x= 1 \\ 12 & ,x= 2 \end{matrix} \right..Then f(x)f(x) is continuous on the set

Cancelling |x − 1| against (x − 1)

x−1∣x−1∣\dfrac{x - 1}{|x - 1|} is not 11; it is ±1\pm 1 depending on the side. Cancelling as if the modulus were absent loses the jump and produces 'continuous on R\mathbb{R}', which is always an option.

Concept 3 of 5: Piecewise with [x] and |x|: Check Every Join and Both Endpoints

A function built from [x][x] and ∣x∣|x| on several intervals can fail at the seams between pieces AND at the integers inside each piece. List every candidate, test each with left/right/value, and count only the failures.

Definition

  • Candidates: every seam between pieces, every integer inside a piece that contains [x][x], and both endpoints of a closed domain.
  • At each candidate compute left limit, right limit and value using the correct piece for each side.
  • Endpoints: at a closed left endpoint only the right limit exists, so continuity there means right limit == value; symmetrically at a closed right endpoint. x+[x]x + [x] at the right endpoint 33 has left limit 3+2=53 + 2 = 5 but value 66 — discontinuous.
  • Two pieces that happen to agree at a seam (x+∣x∣x + |x| and x+[x]x + [x] both give 44 at 22) make that seam continuous even though the formulas differ.
  • Report the count or the set, as asked; 'only three points' means exactly three.

Continuity at a closed endpoint

left endpoint a: lim⁡x→a+f(x)=f(a)right endpoint b: lim⁡x→b−f(x)=f(b)\text{left endpoint } a:\ \lim_{x\to a^+} f(x) = f(a) \qquad \text{right endpoint } b:\ \lim_{x\to b^-} f(x) = f(b)

Worked example

f(x)=[x]+∣x∣f(x) = [x] + |x| on [−1,2)[-1, 2). At how many points is ff discontinuous?
Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q107Hard

Example 3 · Limits · Discontinuities of [x], |x| and sgn x — Counting the Points

Let f:[−1,3]→Rf:[-1,3]\to\mathbb{R} be defined as f(x)={∣x∣+[x],−1≤x<1x+∣x∣,1≤x<2x+[x],2≤x≤3f(x) = \begin{cases}|x|+[x],&-1\leq x<1\\x+|x|,&1\leq x<2\\x+[x],&2\leq x\leq3\end{cases} where [t][t] denotes the greatest integer function. Then ff is discontinuous at

Forgetting the closed right endpoint

On [−1,3][-1, 3] with f(x)=x+[x]f(x) = x + [x] near 33, the value f(3)=6f(3) = 6 is not the left limit 55. A student who only checks interior seams reports two points; the answer is three.

Concept 4 of 5: Composites of [x]: Where Does the Inner Function Cross an Integer?

[g(x)][g(x)] jumps whenever g(x)g(x) crosses an integer. For [x2][x^2] that happens at x=±nx = \pm\sqrt{n}; for [x[x2]]\left[x\left[\frac{x}{2}\right]\right] you have to track both the inner staircase and the product it feeds into.

Definition

  • [g(x)][g(x)] is discontinuous at each xx where g(x)g(x) passes through an integer value, provided gg is continuous and strictly monotone there. Where gg merely touches an integer without crossing, check separately.
  • [x2][x^2] on (0,2)(0, 2): x2x^2 crosses 1,2,31, 2, 3 at x=1,2,3x = 1, \sqrt2, \sqrt3 — three jumps.
  • [−x2][-x^2] uses [−u]=−[u]−1[-u] = -[u] - 1 for non-integer uu: as x2→9−x^2 \to 9^-, [x2]=8[x^2] = 8 and [−x2]=−9[-x^2] = -9; as x2→9+x^2 \to 9^+, [x2]=9[x^2] = 9 and [−x2]=−10[-x^2] = -10. So [x2]−[−x2][x^2] - [-x^2] has left limit 1717 and right limit 1919 at x=3x = 3 — no kk makes it continuous.
  • [x[x2]]\left[x\left[\frac{x}{2}\right]\right] on (−10,10)(-10, 10): the inner [x2]\left[\frac{x}{2}\right] is constant on each [2m,2m+2)[2m, 2m + 2), so the product is mxmx there, and [mx][mx] jumps at multiples of 1∣m∣\frac{1}{|m|} inside that block; the seams x=2mx = 2m must also be tested. The bank's answer is 88 points.

Jumps of a composite

[g(x)] jumps where g(x)∈Z and g crosses it[−u]=−[u]−1 (u∉Z)[g(x)] \text{ jumps where } g(x) \in \mathbb{Z} \text{ and } g \text{ crosses it} \qquad [-u] = -[u] - 1 \ (u \notin \mathbb{Z})

Worked example

How many points of discontinuity does f(x)=[x2]f(x) = [x^2] have on (0,2)(0, 2)?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q147Hard

Example 4 · Limits · Discontinuities of [x], |x| and sgn x — Counting the Points

Let f(x)=[x[x2]]f(x) = \left[x\left[\frac{x}{2}\right]\right], for −10<x<10-10 < x < 10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to

Answering the 'find k' question when no k exists

[x2]−[−x2][x^2] - [-x^2] at 33 has one-sided limits 1717 and 1919. The exam key takes the right-hand value 1919; mathematically no kk works. On the paper choose 1919; in your notes, know why the question is flawed.

Concept 5 of 5: When the Other Factor Vanishes at the Jump

[x][x] jumps by 11 at nn. Multiply it by something that is 00 at nn and continuous there, and the jump is multiplied by 00: both one-sided limits become 00, the value is 00, and the product is continuous.

Definition

  • If gg is continuous at nn with g(n)=0g(n) = 0, and hh is bounded near nn (any [⋅][\cdot] or sign function is), then g⋅h→0=g(n)h(n)g\cdot h \to 0 = g(n)h(n): the product is continuous at nn.
  • [x]sin⁡πx[x]\sin\pi x: sin⁡πn=0\sin\pi n = 0 at every integer, so the product is continuous on all of R\mathbb{R}.
  • [x]cos⁡(2x−1)π2[x]\cos\dfrac{(2x - 1)\pi}{2}: at x=nx = n, cos⁡(2n−1)π2=0\cos\dfrac{(2n - 1)\pi}{2} = 0 — the cosine of an odd multiple of π2\dfrac{\pi}{2} — so this too is continuous at every integer, hence everywhere.
  • If g(n)≠0g(n) \neq 0 the jump survives: [x](x−1)(x−2)[x](x - 1)(x - 2) is discontinuous at every integer except 11 and 22.
  • The test is at each integer separately: which integers make the other factor zero?

A zero swallows a bounded jump

g continuous at n, g(n)=0, ∣h∣≤M near n ⇒ lim⁡x→ng(x)h(x)=0=g(n)h(n)g \text{ continuous at } n,\ g(n) = 0,\ |h| \le M \text{ near } n \ \Rightarrow\ \lim_{x\to n} g(x)h(x) = 0 = g(n)h(n)

Worked example

Where is f(x)=[x]sin⁡πxf(x) = [x]\sin\pi x discontinuous?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q119Moderate

Example 5 · Limits · Discontinuities of [x], |x| and sgn x — Counting the Points

The function f(x)=[x]⋅cos⁡(2x−12π)f(x)=[x]\cdot\cos\left(\dfrac{2x-1}{2}\pi\right), where [⋅][\cdot] denotes the greatest integer function, is discontinuous at

The exam key that contradicts the mathematics

For [x]cos⁡(2x−1)π2[x]\cos\dfrac{(2x - 1)\pi}{2} the official MHT-CET key marks 'all integer points', reasoning only that [x][x] jumps there. The cosine is 00 at every integer, so the function is in fact continuous everywhere. Know the correct mathematics; on that particular paper, the marked answer was the key's.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

  • [x] Is Discontinuous at Every Integer — Counting Them

    Jumps of the greatest-integer function

    lim⁡x→n−[x]=n−1≠n=[n](n∈Z)[kx] jumps at x=mk, m∈Z\lim_{x\to n^-}[x] = n - 1 \ne n = [n] \quad (n \in \mathbb{Z}) \qquad [kx] \text{ jumps at } x = \frac{m}{k},\ m \in \mathbb{Z}
  • Signum-Type Jumps: (x − a)/|x − a| and Products with It

    The sign factor

    x−a∣x−a∣={1,x>a−1,x<ajump of size 2 at a\frac{x - a}{|x - a|} = \begin{cases} 1, & x > a \\ -1, & x < a \end{cases} \qquad \text{jump of size } 2 \text{ at } a
  • Piecewise with [x] and |x|: Check Every Join and Both Endpoints

    Continuity at a closed endpoint

    left endpoint a: lim⁡x→a+f(x)=f(a)right endpoint b: lim⁡x→b−f(x)=f(b)\text{left endpoint } a:\ \lim_{x\to a^+} f(x) = f(a) \qquad \text{right endpoint } b:\ \lim_{x\to b^-} f(x) = f(b)
  • Composites of [x]: Where Does the Inner Function Cross an Integer?

    Jumps of a composite

    [g(x)] jumps where g(x)∈Z and g crosses it[−u]=−[u]−1 (u∉Z)[g(x)] \text{ jumps where } g(x) \in \mathbb{Z} \text{ and } g \text{ crosses it} \qquad [-u] = -[u] - 1 \ (u \notin \mathbb{Z})
  • When the Other Factor Vanishes at the Jump

    A zero swallows a bounded jump

    g continuous at n, g(n)=0, ∣h∣≤M near n ⇒ lim⁡x→ng(x)h(x)=0=g(n)h(n)g \text{ continuous at } n,\ g(n) = 0,\ |h| \le M \text{ near } n \ \Rightarrow\ \lim_{x\to n} g(x)h(x) = 0 = g(n)h(n)

Watch out for (5)

Test yourself on Limits

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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