MHT-CET Maths · Limits

Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

A piecewise function can only fail at the points where its formula changes — so continuity is one equation per junction, and two unknowns need two junctions.

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Why this matters

24 PYQs at 50% HARD, and the most mechanical page in the chapter once the habit is fixed: find the junctions, write left = right = value at each, solve. In the 2021–2025 papers the same three-piece trigonometric function was set four times with the question changed only in what combination of a and b it asks for; the 1 − cos 4x family five times. The HARD tag here comes from junctions hidden inside an inequality or a limit that must be evaluated with different tools on the two sides — never from new theory.

Concept 1 of 5: One Junction: Left Limit = Right Limit = Value

Two formulas glued at

x = c

agree everywhere except possibly at the seam. Continuity at the seam is a single equation: the formula on the left, evaluated at

c

, equals the formula on the right, evaluated at

c

Definition

A piecewise function is continuous away from its junctions automatically (each piece is a nice formula on an open interval). Only the junctions need checking.
At a junction $c$ : compute $\lim_{x\to c^-}$ from the left piece, $\lim_{x\to c^+}$ from the right piece, and $f(c)$ from whichever piece's inequality includes $c$ . Set all three equal.
With one unknown, one junction gives one equation — solve it.
When each piece is a polynomial or a trigonometric function, the one-sided limit is just substitution into that piece.
Check the domain: 'continuous on its domain' or 'on $[-2, 2]$ ' means every junction inside that domain must be tested.

Junction condition

f(x) = \begin{cases} g(x), & x \le c \\ h(x), & x > c \end{cases} \text{ continuous at } c \iff g(c) = \lim_{x\to c^+} h(x)

Worked example

f(x) = 2x + k

for

x \le 1

and

f(x) = x^2 + 3

for

x > 1

. Find

k

f

is continuous.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q122Moderate

Example 1 · Limits · Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

f(x) =\left\{ \begin{matrix} mx+ 1, & x\leqslant \frac{\pi}{2} \\ \sin x+ n, & x>\frac{\pi}{2} \end{matrix} \right.

, is continuous at

x=\frac{\pi}{2},(\text{ }m,n \in\mathbb{Z})

then

Which piece owns the point?

f(c)

comes from the piece whose inequality includes

c

— the one written with

\le

\ge

. When a stem gives a separate value at

c

(

f(3) = a + b

), that value is a THIRD quantity and must equal both one-sided limits.

Drill 1 more on one junction: left limit = right limit = value

Concept 2 of 5: Two Different Formulas Meeting at 0: Compute Each Side with Its Own Tool

The paper's favourite: a trigonometric standard form on the left of

0

, a surd on the right, and a constant

\alpha

0

. The two sides need two different tools — and both must come out equal to

\alpha

Definition

Left of $0$ : typically $\dfrac{1 - \cos kx}{x^2} \to \dfrac{k^2}{2}$ (for $k = 4$ : $8$ ), or $\dfrac{\sin ax}{x} + 3 \to a + 3$ .
Right of $0$ : typically a surd — $\dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$ : rationalise to $\sqrt{16 + \sqrt{x}} + 4 \to 8$ ; or $\dfrac{\sqrt{1 + mx} - \sqrt{1 - mx}}{x} \to m$ .
If both sides are pure limits, they must agree with each other; $\alpha$ is then their common value. If one side is a formula that can be substituted ( $\dfrac{2x + 1}{x - 2} \to -\dfrac{1}{2}$ ), that value is the target for the other side.
When the unknown sits inside the left piece ( $k$ in $1 - \cos kx$ ) and the right piece is fixed, solve $\dfrac{k^2}{2} = 8$ , and read the sign from the options.

The two recurring one-sided limits

\lim_{x\to 0^-}\frac{1 - \cos 4x}{x^2} = 8 \qquad \lim_{x\to 0^+}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} = \lim_{x\to 0^+}\left(\sqrt{16 + \sqrt{x}} + 4\right) = 8

Worked example

f(x) = \dfrac{\sin 3x}{x}

for

x < 0

f(0) = \alpha

f(x) = \dfrac{e^{3x} - 1}{x}

for

x > 0

. Find

\alpha

f

is continuous at

0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q120Moderate

Example 2 · Limits · Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

Given

f(x)=\begin{cases}\dfrac{1-\cos4x}{x^2}, & x<0\\ \alpha, & x=0\\ \dfrac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0\end{cases}

. If

f(x)

is continuous at

x=0

, then value of

\alpha

Substituting into the surd piece

\dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}

\dfrac{0}{0}

0

, not

0

. Multiply by the conjugate

\sqrt{16 + \sqrt{x}} + 4

; the

\sqrt{x}

cancels and the value is

8

. Answering

0

-8

(a sign slip in the conjugate) are the two distractors.

Drill 4 more on two different formulas meeting at 0: compute each side with its own tool

Concept 3 of 5: Exponential Junctions and the Given Value at 0

When a piece is

\dfrac{3\sin x + 5\tan x}{a^{x} - 1}

, its limit is

\dfrac{8}{\log a}

— a formula in the unknown base. Equating it to the value given at

0

turns the unknown into a power of

2

Definition

$\dfrac{p\sin x + q\tan x}{a^{x} - 1} \to \dfrac{p + q}{\log a}$ ; $\dfrac{px + qx\cos x}{b^{x} - 1} \to \dfrac{p + q}{\log b}$ .
Equate to the given $f(0)$ , say $\dfrac{2}{\log 2}$ : $\dfrac{8}{\log a} = \dfrac{2}{\log 2} \Rightarrow \log a = 4\log 2 = \log 16 \Rightarrow a = 16$ .
$\dfrac{8^{x} - 4^{x} - 2^{x} + 1}{x^2} = \dfrac{(4^{x} - 1)(2^{x} - 1)}{x^2} \to \log 4\log 2 = 2(\log 2)^2$ on one side; a polynomial-plus-constant piece $e^{x}\sin x + x + \lambda\log 4$ on the other simply substitutes to $\lambda\log 4 = 2\lambda\log 2$ .
The answer is usually requested as $e^{\lambda}$ or $500e^{\lambda}$ : with $\lambda = \log 2$ , $e^{\lambda} = 2$ .

Exponential-denominator junction

\lim_{x\to 0}\frac{p\sin x + q\tan x}{a^{x} - 1} = \frac{p + q}{\log a} \qquad \log a = k\log 2 \iff a = 2^{k}

Worked example

f(x) = \dfrac{a^{x} - 1}{x}

for

x < 0

f(0) = 2

f(x) = \dfrac{e^{2x} - 1}{x}

for

x > 0

. Find

a

f

is continuous at

0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q146Hard

Example 3 · Limits · Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

f(x)

is continuous at point

x= 0

where

f(x) =\left\{ \begin{matrix} \frac{3\sin x+ 5\tan x}{a^{x}- 1} & ,x< 0 \\ \frac{2}{\log2} & ,x= 0 \\ \frac{8x+ 2xcosx}{{\text{ }b}^{x}- 1} & ,x> 0 \end{matrix} \right.

then the values of

a

and

b

, respectively, are .

log a = 4 log 2 means a = 16, not a = 8

4\log 2 = \log 2^4 = \log 16

. The option

a = 8

comes from

2 \times 4

; the exponent rule for logs is what the stem is testing.

Drill 2 more on exponential junctions and the given value at 0

Concept 4 of 5: Two Junctions, Two Unknowns: Set Up a Linear System

Three pieces have two seams, and each seam gives one equation in

a

and

b

. Write both, solve the pair, then compute whatever combination the stem asks for —

a - b

2a + 3b

, the ordered pair.

Definition

Count the junctions first. Three pieces → two junctions → two equations. Unknowns beyond that number cannot be determined by continuity alone.
The recurring paper: $x + a\sqrt2\sin x$ on $[0, \frac{\pi}{4}]$ , $2x\cot x + b$ on $[\frac{\pi}{4}, \frac{\pi}{2}]$ , $a\cos 2x - b\sin x$ on $(\frac{\pi}{2}, \pi]$ . At $\frac{\pi}{4}$ : $\frac{\pi}{4} + a = \frac{\pi}{2} + b$ , i.e. $a - b = \frac{\pi}{4}$ . At $\frac{\pi}{2}$ : $2\cdot\frac{\pi}{2}\cdot 0 + b = -a - b$ , i.e. $a + 2b = 0$ . Hence $a = \frac{\pi}{6}$ , $b = -\frac{\pi}{12}$ .
Junctions can be hidden in an inequality: $|2x - 3| \ge 2$ means $x \le \frac12$ or $x \ge \frac52$ , so the seams are at $\frac12$ and $\frac52$ . Solve the inequality before writing anything.
A seam that involves a limit rather than substitution ( $\dfrac{\sin ax}{x} + 3$ at $0$ ) is handled with the standard form, then the equation is linear as usual.
Pieces like $\dfrac{2x^2}{a}$ and $a$ meeting at $1$ give $a^2 = 2$ — a quadratic; carry both roots to the second junction and let it decide.

The two-seam system

\text{seam } c_1:\ g(c_1) = h(c_1) \qquad \text{seam } c_2:\ h(c_2) = k(c_2) \qquad \Rightarrow\ \text{two linear equations in } a, b

Worked example

f(x) = x + a

for

x \le 0

2x + b

for

0 < x \le 1

3

for

x > 1

. Find

a

and

b

f

is continuous.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q120Hard

Example 4 · Limits · Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

The values of a and b, so that the function

f(x)=\begin{cases}x+a\sqrt{2}\sin x,&0\leq x\leq\frac{\pi}{4}\\2x\cot x+b,&\frac{\pi}{4}\leq x\leq\frac{\pi}{2}\\a\cos2x-b\sin x,&\frac{\pi}{2}<x\leq\pi\end{cases}

is continuous for

0\leq x\leq\pi

, are respectively given by

Applying continuity at only one seam

With two unknowns, one equation leaves a free parameter, and any option can be 'reached'. The 2022 sitting's own key was wrong for exactly this reason — it used one seam and printed

\frac{13}{5}

where two seams give

\frac{23}{5}

. Count the seams before solving.

Drill 7 more on two junctions, two unknowns: set up a linear system

Concept 5 of 5: The Squeeze: x² sin(1/x) Is Continuous for Any Coefficient

\sin\dfrac{1}{x}

oscillates wildly near

0

and has no limit — but multiply it by

x^2

and the oscillation is crushed between

-x^2

and

x^2

, both of which go to

0

. The product is continuous at

0

no matter what constant sits in front.

Definition

Squeeze theorem: if $g(x) \le f(x) \le h(x)$ near $c$ and $g, h \to L$ , then $f \to L$ .
$|\sin u| \le 1$ for every $u$ , so $\left|x^2\sin\dfrac{1}{x}\right| \le x^2 \to 0$ and $\left|x\sin\dfrac{1}{x}\right| \le |x| \to 0$ .
Consequence: $f(x) = bx^2\sin\dfrac{1}{x}$ for $x > 0$ , $f(0) = 0$ , is continuous at $0$ for every real $b$ — the parameter is unconstrained.
Likewise $a^2(x - |x|)$ for $x < 0$ equals $2a^2x \to 0$ for every $a$ . A stem asking 'for which $a, b$ is $f$ continuous' can have the answer 'all real $a$ and $b$ '.
Without the crushing factor, $\sin\dfrac{1}{x}$ alone has no limit at $0$ and no choice of $f(0)$ makes it continuous.

Squeeze at 0

\left|x^2\sin\frac{1}{x}\right| \le x^2 \to 0 \ \Rightarrow\ \lim_{x\to 0} x^2\sin\frac{1}{x} = 0

Worked example

f(x) = x\sin\dfrac{1}{x}

for

x \neq 0

and

f(0) = 0

. Is

f

continuous at

0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q108Moderate

Example 5 · Limits · Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

f(x) = \begin{cases} a^2(x-|x|), & x < 0 \\ 0, & x = 0 \\ bx^2\sin\frac{1}{x}, & x > 0 \end{cases}

is continuous at

x = 0

, then

Looking for a constraint that is not there

When both pieces tend to

0

for every value of the parameters, the answer is 'any real

a

, any real

b

'. Options that restrict

a

to rationals or irrationals are noise — nothing in the limit distinguishes them.

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)

One Junction: Left Limit = Right Limit = Value
Junction condition
$f(x) = \begin{cases} g(x), & x \le c \\ h(x), & x > c \end{cases} \text{ continuous at } c \iff g(c) = \lim_{x\to c^+} h(x)$
Two Different Formulas Meeting at 0: Compute Each Side with Its Own Tool
The two recurring one-sided limits
$\lim_{x\to 0^-}\frac{1 - \cos 4x}{x^2} = 8 \qquad \lim_{x\to 0^+}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} = \lim_{x\to 0^+}\left(\sqrt{16 + \sqrt{x}} + 4\right) = 8$
Exponential Junctions and the Given Value at 0
Exponential-denominator junction
$\lim_{x\to 0}\frac{p\sin x + q\tan x}{a^{x} - 1} = \frac{p + q}{\log a} \qquad \log a = k\log 2 \iff a = 2^{k}$
Two Junctions, Two Unknowns: Set Up a Linear System
The two-seam system
$\text{seam } c_1:\ g(c_1) = h(c_1) \qquad \text{seam } c_2:\ h(c_2) = k(c_2) \qquad \Rightarrow\ \text{two linear equations in } a, b$
The Squeeze: x² sin(1/x) Is Continuous for Any Coefficient
Squeeze at 0
$\left|x^2\sin\frac{1}{x}\right| \le x^2 \to 0 \ \Rightarrow\ \lim_{x\to 0} x^2\sin\frac{1}{x} = 0$

Watch out for (5)

Which piece owns the point?
→ One Junction: Left Limit = Right Limit = Value
Substituting into the surd piece
→ Two Different Formulas Meeting at 0: Compute Each Side with Its Own Tool
log a = 4 log 2 means a = 16, not a = 8
→ Exponential Junctions and the Given Value at 0
Applying continuity at only one seam
→ Two Junctions, Two Unknowns: Set Up a Linear System
Looking for a constraint that is not there
→ The Squeeze: x² sin(1/x) Is Continuous for Any Coefficient

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Related notes

Algebraic Limits — rationalisation for the surd pieces →