Q.Prove that the greatest integer function defined by f(x)=[x], 0<x<3, is not differentiable at x=1 and x=2.
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Continuity of the Greatest Integer Function
The greatest integer function f(x)=⌊x⌋ returns the largest integer not exceeding x: ⌊2.3⌋=2, ⌊−1.2⌋=−2, ⌊4⌋=4. Its graph is a staircase — flat segments that jump up by 1 at every integer.
The intuition
Walk along the graph from left to right. Near a non-integer such as x=1.5 the function is flat at 1; nudge x a little either way and the value does not change, so nothing is broken there. But as you approach an integer like x=2 from the left the value is stuck at 1, and the instant you reach x=2 it leaps to 2. That sudden leap is a break.
⌊x⌋ is continuous at every non-integer and discontinuous at every integer.
Why integers fail
At an integer n the one-sided limits disagree:
limx→n−⌊x⌋=n−1,limx→n+⌊x⌋=n,⌊n⌋=n.
Since the left- and right-hand limits differ, limx→n⌊x⌋ does not exist, so continuity fails. This is a jump discontinuity, and the jump is always exactly 1. At a non-integer c there is a whole small interval on which f is constant equal to ⌊c⌋, so the limit exists and matches f(c) — the function is continuous.
How to test it …
A function that is not continuous at a point cannot be differentiable there. The greatest integer function f(x)=[x] jumps at each integer.
At x=1: limx→1−[x]=0 but limx→1+[x]=1, so the limit does not exist and f is discontinuous — hence not differentiable. Checking derivatives with f(1)=1: the right-hand derivative limh→0+h[1+h]−1=limh→0+h0=0, while the left-hand derivative limh→0−h[1+h]−1=limh→0−h−1→+∞; the …
[x] has a jump at each integer, so it is discontinuous — and therefore not differentiable — at x=1 and x=2; the one-sided derivatives there also disagree.
On 0<x<3 the greatest integer function is a staircase: [x]=0 on (0,1), [x]=1 on [1,2), [x]=2 on [2,3). Differentiability requires continuity first, and where the graph jumps it cannot be continuous.
Discontinuity forces non-differentiability at x=1
Left: for x just below 1, [x]=0, so limx→1−[x]=0.
Right: for x just above 1, [x]=1, so limx→1+[x]=1.
The one-sided limits differ, so limx→1[x] does not exist — f is discontinuous, hence not differentiable at x=1.
Confirm with the derivative definition at x=1
Using f(1)=[1]=1:
f+′(1)=limh→0+h[1+h]−1=limh→0+h1−1=0,
f−′(1)=limh→0−h[1+h]−1=limh→0−h0−1=limh→0−h−1→+∞.
The right-hand derivative is 0 and the left-hand derivative diverges, so they are unequal — f′(1) does not exist. …
Method: Showing Non-Differentiability at a Jump Discontinuity
Use this method for functions like the greatest integer (floor) function that jump abruptly at certain points — this route is shorter than the corner-point method above because differentiability can be ruled out immediately once a jump is shown.
Steps
Step 1: Recall that differentiability requires continuity first
If a function is not even continuous at a point, it cannot be differentiable there — there is no need to compute a derivative limit at all. This shortcut saves work whenever a jump can be shown directly.
Step 2: Compute the left-hand and right-hand limits of the function itself (not yet the derivative) at the point in question
For the greatest integer function [x] at an integer n, evaluate [x] for x slightly less than n and slightly greater than n separately.
Step 3: Compare the two one-sided limits to the function's actual value
If limx→n−f(x)=limx→n+f(x), the two-sided limit does not exist, so f is discontinuous at n — and, by Step 1's logic, therefore automatically not differentiable there. …
Common Mistakes
Mistake 1: Trying to prove non-differentiability directly from the derivative limit without first checking continuity
Why it's wrong: it's more work, and easy to make sign errors, to jump straight into computing limh→0h[n+h]−n from both sides without first noticing the simpler fact that [x] isn't even continuous at n. Correct approach: always check continuity first at a suspected trouble point — if it fails, non-differentiability follows immediately and no derivative computation is required.
Mistake 2: Evaluating [1+h] or [2+h] incorrectly for negative h
Why it's wrong: for h a small negative number, 1+h is just below 1 (e.g. 0.99), so [1+h]=0, not 1 — students sometimes assume the floor value doesn't change until h crosses a whole unit. Correct approach: pick a concrete small value (like h=−0.01) and evaluate [1+h] numerically before generalizing. …
- CBSE 2025Set ANNUAL1 markMCQQ.The function f(x)=[x], where [x] denotes the greatest integer less than or equal to x, is continuous on(a) x=1(b) x=1.5(c) x=−2(d) x=4
›Reveal solutionSolution
[x] jumps at every integer, so it's continuous exactly at the non-integer points.
The greatest integer function f(x)=[x] has a jump discontinuity at every integer n, because limx→n−[x]=n−1 while limx→n+[x]=n=f(n) — the left and right limits disagree.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The function f(x)=[x], where [x] denotes the greatest integer function, is continuous at:(a) 4(b) −2(c) 1(d) 1.5
›Reveal solutionSolution
The greatest integer function [x] is discontinuous at every integer and continuous everywhere else.
At an integer n, limx→n−[x]=n−1 while limx→n+[x]=n=[n], so the left and right limits differ — [x] is discontinuous at every integer. Among the options, 4,−2,1 are integers (discontinuous), while 1.5 is not …
- CBSE 2025Set ANNUAL1 markMCQQ.At x = 2, f(x) = [x] (greatest integer function) is –(i) continuous but not differentiable(ii) differentiable but not continuous(iii) continuous as well as differentiable(iv) neither continuous nor differentiable
›Reveal solutionSolution
The greatest integer function jumps at every integer, so it fails continuity (and hence differentiability) at x = 2.
For f(x)=[x] (greatest integer function), at an integer n:
limx→n−f(x)=n−1 while limx→n+f(x)=n.
At x=2: left-hand limit =1, right-hand limit =2=f(2). Since the left and right limits differ, limx→2f(x) does not exist, so f is discontinuous at x=2.
…
- CBSE 2024Set A11 markMCQQ.The function f:R→R defined as f(x)=[x], where [x] denotes the greatest integer less than or equal to x. For what values of x in the interval 2<x<5 given below f(x) is not differentiable?(a) 2 and 5(b) 3 and 5(c) 4 and 5(d) 3 and 4
›Reveal solutionSolution
[x] jumps at every integer, and the only integers strictly inside (2,5) are 3 and 4, so (d).
The step function f(x)=[x] is constant between consecutive integers and jumps at each integer, so it fails to be continuous (hence not d …
- CBSE 2020Set 65/1/11 markQ.The greatest integer function f(x)=[x], defined for 0<x<2, is not differentiable at x=__________.
›Reveal solutionSolution
The greatest integer function [x] on (0,2) has integer jumps at x=1 where left and right derivatives differ, so it is not differentiable at x=1.
The greatest integer function f(x)=[x] returns the largest integer less than or equal to x. On the open interval (0,2), the only integer inside is 1. At every other point, the function is locally constant — flat horizontal segments — so the derivative exists and equals 0. The trouble is only at the jump.
For differentiability at a point, the function must be continuous there first. But [x] has a jump discontinuity at every integer: the left-hand limit and right-hand limit differ by 1. At x=1, the left limit is 0 and the right limit is 1, so the function isn't even continuous — and therefore cannot be differentiable.
Even if we ignored continuity and tried to compute the derivative from the definition, the left and right difference quotients would give different results. Let's check that explicitly.
- Left-hand derivative at x=1 For x just less than 1, say x=1−h with h>0 small, [x]=0. The difference quotient is
−hf(1−h)−f(1)=−h0−1=−h−1=h1.
As h→0+, this blows up to +∞. So the left-hand derivative does not exist as a finite number.
- Right-hand derivative at x=1 For x just greater than 1, say x=1+h with h>0, [x]=1. The difference quotient is hf(1+h)−f(1)=h1−1=0. …
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