MathsISC Class 12 · Continuity, Differentiability and Differentiation

Continuity and Differentiability of Piecewise Functions

Updated 15 October 2026 · Fact-checked

A function is continuous at x = a when the left-hand limit, the right-hand limit and f(a) all exist and are equal. It is differentiable at a when the left-hand and right-hand derivatives are also equal. For piecewise functions, test each joining point separately, using the correct piece on each side.

Understand Continuity and Differentiability of Piecewise Functions

A piecewise function uses different formulas on different intervals, such as
$f (x) = ⎩ ⎨ ⎧ x + 2, - x^{2} - 2, x, x < 0 0 \leq x < 1 x \geq 1$
Each formula on its own is a polynomial or another standard function, which is continuous wherever it is defined. So the only places where trouble can occur are the joining points (here $x = 0$ and $x = 1$ ) and any points where a formula is undefined.

Continuity at a point. $f$ is continuous at $x = a$ if
$lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a) .$
Three things must hold: the left-hand limit (LHL) exists, the right-hand limit (RHL) exists, and both equal the value of the function. For LHL use the piece that applies just to the left of $a$ ; for RHL, the piece just to the right; for $f (a)$ , the piece whose interval contains $a$ itself (look for $\leq$ or $\geq$ ). The ISC 2025 examiners noted that candidates used the wrong pieces for the one-sided limits. In the example above, at $x = 0$ : LHL $= 2$ , RHL $= - 2$ , so $f$ is discontinuous; at $x = 1$ : LHL $= - 3$ , RHL $= 1$ , discontinuous again. That gives 2 points of discontinuity.

Types of discontinuity.

Removable: the limit exists but is not equal to $f (a)$ (or $f (a)$ is not defined). Redefining $f (a)$ as the limit removes it. Example: $\frac{s i n 3 x}{x}$ at $0$ has limit 3.
Jump (non-removable): LHL and RHL both exist but differ, as for the greatest integer function $[x]$ at every integer.

"Find k" questions. If $f (x) = \frac{expression}{x - a}$ for $x \neq = a$ and $f (a) = k$ , the expression usually has $(x - a)$ as a factor. Factorise, cancel (allowed because $x \neq = a$ in the limit), substitute $x = a$ , and set the limit equal to $k$ . For two unknowns, write one equation from each side of the joining point.

Differentiability. $f$ is differentiable at $a$ if the left-hand derivative
$L f^{'} (a) = lim_{h \to 0^{-}} \frac{f ( a + h ) - f ( a )}{h}$
equals the right-hand derivative $R f^{'} (a)$ (the same limit with $h \to 0^{+}$ ). For a piecewise function that is continuous at the joining point, you can find these by differentiating each piece and substituting $a$ . Geometrically, a differentiable function has a single tangent: no corner, no jump and no vertical tangent. The ISC 2024 paper asked candidates to read points of non-differentiability from a graph; the corners and breaks are exactly those points.

The key link. Differentiable $\Rightarrow$ continuous, but not the other way round. $∣ x ∣$ is continuous at 0 but has a corner there ( $L f^{'} (0) = - 1$ , $R f^{'} (0) = 1$ ), so it is not differentiable. On the other hand $x ∣ x ∣$ is differentiable at 0 with derivative 0. So if a function is discontinuous at a point, it is automatically not differentiable there; if it is continuous, you must still check the one-sided derivatives.

Standard functions. Polynomials, $sin x$ , $cos x$ , $e^{x}$ are continuous everywhere; $lo g x$ for $x > 0$ ; rational functions except where the denominator is 0; $∣ x ∣$ everywhere (but not differentiable at 0); $[x]$ everywhere except at integers. Sums, differences, products and compositions of continuous functions are continuous; quotients are continuous where the denominator is non-zero.

Continuity and Differentiability of Piecewise Functions: formulas to remember

Continuity at a: $lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)$; All three must exist and be equal.
Left-hand derivative: $L f^{'} (a) = lim_{h \to 0^{-}} \frac{f ( a + h ) - f ( a )}{h}$; Use the piece to the left of a.
Right-hand derivative: $R f^{'} (a) = lim_{h \to 0^{+}} \frac{f ( a + h ) - f ( a )}{h}$; Differentiable at a if Lf′(a) = Rf′(a).
Link: $differentiable \Rightarrow continuous$; The converse is false: |x| at 0.
Standard limit: $lim_{x \to 0} \frac{s i n x}{x} = 1, lim_{x \to 0} \frac{1 - c o s x}{x ^{2}} = \frac{1}{2}$; Used in many "find k" questions.
Factor limit: $lim_{x \to a} \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}$; Or factorise and cancel (x − a).
Greatest integer function: $[x] is discontinuous at every integer$; A jump of 1 at each integer; continuous elsewhere.

How to solve Continuity and Differentiability of Piecewise Functions questions

Apply these steps at each joining point of a piecewise function.

1List the joining points and any points where a piece is undefined. Away from them, the function is continuous because each piece is.
2At each joining point $a$ , write three lines: LHL (left piece), RHL (right piece), and $f (a)$ (the piece with $\leq$ or $\geq$ at $a$ ).
3Evaluate each limit by substitution; if you get $\frac{0}{0}$ , factorise and cancel, or use a standard limit.
4Conclude: continuous if all three are equal; removable discontinuity if the limit exists but differs from $f (a)$ ; jump if LHL ≠ RHL.
5For "find $k$ " (or $a$ and $b$ ), set LHL $= f (a) =$ RHL and solve the resulting equations.
6For differentiability, first confirm continuity. Then differentiate each piece and compare the left and right derivatives at $a$ .
7State the final answer for every joining point, for example "continuous but not differentiable at $x = 2$ ".

Continuity and Differentiability of Piecewise Functions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Continuous but not differentiable

Suppose an electricity company charges Rs 5 per unit for the first 100 units in a month and Rs 7 per unit for every unit after that. The bill for $u$ units is $E (u) = 5 u$ for $0 \leq u \leq 100$ and $E (u) = 500 + 7 (u - 100)$ for $u > 100$ .

Is $E$ continuous at $u = 100$ ? Is it differentiable there? Find the bill for 150 units.

Try it first, then show the solutionSolution

LHL $= 5 \times 100 = 500$ ; RHL $= 500 + 7 (0) = 500$ ; $E (100) = 500$ . All equal, so $E$ is continuous at 100: the bill does not jump.
Left derivative $= 5$ , right derivative $= 7$ . They differ, so $E$ is not differentiable at 100: the graph has a corner.
Bill for 150 units: $500 + 7 \times 50 = 850$ rupees.

Answer: Continuous but not differentiable at 100 units; the bill for 150 units is Rs 850.

Memory hook: A slab tariff bends but never breaks: continuous, with a corner where the rate changes.

Jump discontinuity

Imagine a taxi that charges a flat Rs 30 for any trip up to 2 km, but for longer trips charges $F (d) = 40 + 12 (d - 2)$ rupees for $d > 2$ km.

Test the continuity of $F$ at $d = 2$ . Is the discontinuity removable? What is the fare for a 5 km trip?

Try it first, then show the solutionSolution

LHL $= 30$ and $F (2) = 30$ ; RHL $= 40 + 12 (0) = 40$ .
LHL ≠ RHL, so $F$ is discontinuous at $d = 2$ , with a jump of Rs 10.
Since the two one-sided limits differ, no choice of $F (2)$ can fix it: the discontinuity is not removable.
Fare for 5 km: $40 + 12 \times 3 = 76$ rupees.

Answer: Jump discontinuity of Rs 10 at 2 km (not removable); a 5 km trip costs Rs 76.

Memory hook: If one more metre costs Rs 10 extra at once, the fare function has a jump.

Tricks to remember Continuity and Differentiability of Piecewise Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Left, right, value
Continuity is a three-way handshake: LHL, RHL and f(a) must all agree.
Use it when: Setting out every continuity answer.
The equals sign decides f(a)
Look for ≤ or ≥: the piece with the equals sign at a gives f(a).
Use it when: Avoiding the wrong-piece error.
Smooth implies joined
Differentiable means smooth; smooth curves are always joined (continuous), but a joined curve can still have a corner.
Use it when: Assertion-reason items on the link.
Corner, cusp, jump
Not differentiable where the graph has a corner, a vertical tangent or a break.
Use it when: Reading graphs, as in ISC 2024 Q1(iv).

Continuity and Differentiability of Piecewise Functions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Find k by factorising · 2 marks · Short answerBoard level

Find $k$ so that $f (x) = ⎩ ⎨ ⎧ \frac{x ^{2} + 2 x - 15}{x - 3}, k, x \neq = 3 x = 3$ is continuous at $x = 3$ .

Try it first, then show the solutionSolution

For continuity, $lim_{x \to 3} f (x) = f (3) = k$ . (½ mark)
$lim_{x \to 3} \frac{( x + 5 ) ( x - 3 )}{x - 3} = lim_{x \to 3} (x + 5) = 8$ . (1 mark)
So $k = 8$ . (½ mark)

Answer: $k = 8$ .

Example 2: Counting points of discontinuity · 1 mark · MCQBoard level

The number of points of discontinuity of $f (x) = ⎩ ⎨ ⎧ 2 x + 1, x^{2} + 2, 3 x + 1, x < 1 1 \leq x < 2 x \geq 2$ is:
(a) 0
(b) 1
(c) 2
(d) 3

Try it first, then show the solutionSolution

At $x = 1$ : LHL $= 3$ , RHL $= 3$ , $f (1) = 3$ : continuous.
At $x = 2$ : LHL $= 6$ , RHL $= 7$ : discontinuous.

Answer: (b) 1.

Example 3: Two unknowns at one joining point · 3 marks · Short answerHarder board level

Find $a$ and $b$ if $f (x) = ⎩ ⎨ ⎧ x^{2} + a, 5, b x - 1, x < 2 x = 2 x > 2$ is continuous at $x = 2$ .

Try it first, then show the solutionSolution

LHL $= 4 + a$ , RHL $= 2 b - 1$ , $f (2) = 5$ . (1 mark)
LHL $= f (2)$ : $4 + a = 5 \Rightarrow a = 1$ . (1 mark)
RHL $= f (2)$ : $2 b - 1 = 5 \Rightarrow b = 3$ . (1 mark)

Answer: $a = 1$ , $b = 3$ .

Example 4: A corner: continuous but not differentiable · 3 marks · Short answerHarder board level

Show that $f (x) = ∣ x - 2∣$ is continuous but not differentiable at $x = 2$ , and that $g (x) = x ∣ x ∣$ is differentiable at $x = 0$ .

Try it first, then show the solutionSolution

For $f$ : LHL $=$ RHL $= 0 = f (2)$ , so $f$ is continuous at 2. (1 mark)
$L f^{'} (2) = lim_{h \to 0^{-}} \frac{∣ h ∣}{h} = - 1$ and $R f^{'} (2) = lim_{h \to 0^{+}} \frac{∣ h ∣}{h} = 1$ . They differ, so $f$ is not differentiable at 2. (1 mark)
For $g$ : $\frac{g ( 0 + h ) - g ( 0 )}{h} = \frac{h ∣ h ∣}{h} = ∣ h ∣ \to 0$ from both sides, so $g^{'} (0) = 0$ . (1 mark)

Answer: $∣ x - 2∣$ has a corner at 2; $x ∣ x ∣$ is differentiable at 0 with derivative 0.

Example 5: Removable or not? · 3 marks · Short answerHarder board level

(i) $f (x) = \frac{s i n 3 x}{x}$ for $x \neq = 0$ and $f (0) = 2$ . Classify the discontinuity at 0 and remove it if possible.
(ii) Is the discontinuity of $g (x) = [x]$ at $x = 1$ removable?

Try it first, then show the solutionSolution

(i) $lim_{x \to 0} \frac{s i n 3 x}{x} = 3 lim_{x \to 0} \frac{s i n 3 x}{3 x} = 3 \neq = 2 = f (0)$ . The limit exists but differs from $f (0)$ : a removable discontinuity. Redefine $f (0) = 3$ . (2 marks)
(ii) LHL $= 0$ and RHL $= 1$ at $x = 1$ . The limit does not exist, so the jump cannot be removed. (1 mark)

Answer: (i) Removable; set $f (0) = 3$ . (ii) Not removable (jump discontinuity).

Example 6: Continuous and differentiable at a joining point · 5 marks · Long answerHOTS / competency

Find $a$ and $b$ so that $f (x) = {a x^{2} + b, 4 x - 1, x < 1 x \geq 1$ is differentiable at $x = 1$ . Find $f^{'} (1)$ .

Try it first, then show the solutionSolution

Differentiability needs continuity first. LHL $= a + b$ , RHL $= f (1) = 3$ . So $a + b = 3$ . (1 mark)
For $x < 1$ , $f^{'} (x) = 2 a x$ , so $L f^{'} (1) = 2 a$ . For $x > 1$ , $f^{'} (x) = 4$ , so $R f^{'} (1) = 4$ . (2 marks)
Equal derivatives: $2 a = 4 \Rightarrow a = 2$ , and then $b = 1$ . (1 mark)
Then $f^{'} (1) = 4$ . (1 mark)

Answer: $a = 2$ , $b = 1$ , and $f^{'} (1) = 4$ .

Where students lose marks in Continuity and Differentiability of Piecewise Functions

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong piece for a one-sided limit
The ISC 2025 examiners noted that candidates used the wrong pieces of the function for LHL and RHL.
Wrong working
At $x = 1$ for $f (x) = - x^{2} - 2$ on $[0, 1)$ and $x$ on $[1, \infty)$ : LHL $= f (1) = 1$ .
Right working
LHL uses the piece on $[0, 1)$ : $- 1 - 2 = - 3$ . RHL $= 1$ . Discontinuous.
Do this: Write the interval next to each limit before substituting.
Usually costs: 1 mark in Section A, more in longer questions.
Mistake: Not evaluating the 0/0 limit
In ISC 2024 few candidates could evaluate the limit in a "find k" question.
Do this: If substitution gives 0/0, factorise and cancel (x − a), or use a standard limit such as sin x/x → 1.
Usually costs: The whole 2-mark question.
Mistake: Assuming continuity implies differentiability
Corners such as |x| at 0 are continuous but not differentiable.
Do this: Always compare the left and right derivatives separately.
Usually costs: 1 mark.
Mistake: Checking derivatives at a discontinuity
Differentiating the pieces at a point where the function jumps can give equal slopes, which is misleading.
Do this: Check continuity first; if the function is not continuous, it is not differentiable and you can stop.
Usually costs: A wrong conclusion, usually 1 to 2 marks.

Exam tips

Write LHL, RHL and $f (a)$ on separate lines with the piece used for each; examiners give a mark for each.
In "find $k$ " questions, show the factorisation or standard limit you used, then write "for continuity, $k = \dots$ ".
For graph-reading questions, look for corners, sharp points and breaks: these are the points of non-differentiability.
If a function is defined with $\leq$ on one side and $<$ on the other, the equals sign tells you which piece gives $f (a)$ .
When two unknowns are involved, you need two conditions: one from continuity and one from differentiability (or from two joining points).

Practice questions on Continuity and Differentiability of Piecewise Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
At $x = 0$ , the function $f (x) = ∣ x ∣$ is:
(a) continuous and differentiable
(b) continuous but not differentiable
(c) differentiable but not continuous
(d) neither continuous nor differentiable
Hint
Compare $L f^{'} (0)$ and $R f^{'} (0)$ .
Answer
(b). LHL $=$ RHL $= f (0) = 0$ , but $L f^{'} (0) = - 1 \neq = 1 = R f^{'} (0)$ .
Q2 · 2 marks · Short answerBoard level
Find $k$ if $f (x) = ⎩ ⎨ ⎧ \frac{1 - cos 2 x}{x ^{2}}, k, x \neq = 0 x = 0$ is continuous at 0.
Hint
$1 - cos 2 x = 2 sin^{2} x$ .
Answer
$lim_{x \to 0} \frac{2 s i n ^{2} x}{x ^{2}} = 2$ , so $k = 2$ .
Q3 · 3 marks · Short answerHarder board level
Find all points of discontinuity of $f (x) = ⎩ ⎨ ⎧ 2 - x, x^{2}, 5 - x, x < 1 1 \leq x < 2 x \geq 2$ .
Hint
Check $x = 1$ and $x = 2$ only.
Answer
At $x = 1$ : LHL $= 1$ , RHL $= 1$ , $f (1) = 1$ : continuous. At $x = 2$ : LHL $= 4$ , RHL $= 3$ : discontinuous. The only point is $x = 2$ .
Q4 · 2 marks · Short answerHarder board level
Where is $f (x) = [x]$ discontinuous on the interval $(0, 3)$ ? Are these discontinuities removable?
Hint
Look at the integers inside the interval.
Answer
At $x = 1$ and $x = 2$ , where LHL and RHL differ by 1. They are jump discontinuities, so not removable.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $f (x) = ∣ x - 3∣$ is continuous at $x = 3$ .
Reason (R): Every function that is continuous at a point is differentiable at that point.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Think of a corner.
Answer
(c). A is true (all three values are 0); R is false, since $∣ x - 3∣$ itself is continuous but not differentiable at 3.
Q6 · 3 marks · Short answerHOTS / competency
Find the points where $f (x) = ∣ x ∣ + ∣ x - 1∣$ is not differentiable, and show that it is continuous everywhere.
Hint
Write $f$ as three pieces.
Answer
$f (x) = 1 - 2 x$ for $x < 0$ , $1$ for $0 \leq x \leq 1$ , $2 x - 1$ for $x > 1$ . The pieces agree at 0 and 1 (value 1), so $f$ is continuous. Derivatives: $- 2$ then $0$ at $x = 0$ ; $0$ then $2$ at $x = 1$ . Not differentiable at $x = 0$ and $x = 1$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a courier charges $C (w)$ rupees for a parcel of $w$ kg: Rs 50 for $0 < w \leq 1$ ; $50 + 30 (w - 1)$ for $1 < w \leq 5$ ; $k + 20 w$ for $w > 5$ .
(i) Show that $C$ is continuous at $w = 1$ . (1 mark)
(ii) Find $k$ so that $C$ is continuous at $w = 5$ . (1 mark)
(iii) With this $k$ , is $C$ differentiable at $w = 5$ ? (1 mark)
(iv) What does your answer to (iii) mean for a customer? (1 mark)
Hint
Compare the per-kg rates on each side of 5 kg.
Answer
(i) LHL $= 50$ , RHL $= 50 + 0 = 50$ , $C (1) = 50$ . (ii) $C (5) = 50 + 120 = 170$ ; RHL $= k + 100$ , so $k = 70$ . (iii) No: left derivative 30, right derivative 20. (iv) The price has no jump at 5 kg, but each extra kilogram beyond 5 kg costs Rs 20 instead of Rs 30.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the condition for continuity at $x = a$ .Show answer
$lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)$ .
2.Is every continuous function differentiable?Show answer
No. For example $∣ x ∣$ is continuous but not differentiable at 0.
3.Evaluate $lim_{x \to 0} \frac{s i n 5 x}{x}$ .Show answer
5.
4.What kind of discontinuity does $[x]$ have at $x = 2$ ?Show answer
A jump (non-removable) discontinuity.
5.Find $L f^{'} (0)$ and $R f^{'} (0)$ for $f (x) = ∣ x ∣$ .Show answer
$- 1$ and $1$ .
6.If $f$ is not continuous at $a$ , can it be differentiable there?Show answer
No; differentiability implies continuity.

Continuity and Differentiability of Piecewise Functions: frequently asked questions

How do I check continuity of a piecewise function?

Find the joining points. At each one, compute the left-hand limit from the left piece, the right-hand limit from the right piece, and the value from the piece that includes the point. If all three agree, the function is continuous there.

What is a removable discontinuity?

It is a point where the limit of the function exists but does not equal the function value, or the function is not defined there. Redefining the value as the limit makes the function continuous.

Why is |x| not differentiable at 0?

Its left-hand derivative is −1 and its right-hand derivative is 1. Because they differ, there is no single tangent at 0; the graph has a corner.

Can I differentiate each piece to check differentiability?

Yes, as long as the function is continuous at the joining point. Differentiate each piece, substitute the joining point and compare. If the function is not continuous there, it is not differentiable, whatever the slopes.

Is the continuity of [x] in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists the continuity and differentiability of functions such as |x| and [x], removable discontinuities and the continuity of a function at a point and in an interval.