Chapter 5 · Continuity and Differentiability

5.1 Continuity

A function is continuous at $c$ when $\lim_{x\to c}f(x)$ exists and equals $f(c)$ — no hole, no jump, no stray point. Sums, products, quotients and composites of continuous functions are continuous.

A thermometer outside your window records the temperature through the day. The reading can rise quickly or fall quickly, but it cannot go from $28^\circ\text{C}$ to $31^\circ\text{C}$ without passing through every temperature in between: the graph of temperature against time has no breaks. Now look at an electricity bill in many states, where the price of a unit changes when you cross a slab such as $100$ units, or at a parking ticket, which jumps from ₹20 to ₹35 the moment you stay a second past one hour. Those graphs do break. Engineers, economists and doctors need to know which kind of quantity they are dealing with: a break may signal a fault in a sensor, a tariff that is unfair at the slab boundary, or a dose that changes abruptly.

In this section we turn “a graph with no breaks” into a precise test that uses the left-hand and right-hand limits you met in Class 11 Mathematics, Ch. 12 — Limits and Derivatives. The test works one point at a time, so it can be applied to piecewise formulas, to the modulus and greatest integer functions, and to functions far too complicated to sketch. It also lets us choose constants so that the pieces of a formula join up — a question the Board asks every year.

Continuity at a Point

Recall from Class 11: the left-hand limit $\lim\limits_{x\to c^-}f(x)$ is the value $f(x)$ approaches as $x$ approaches $c$ through values less than $c$; the right-hand limit $\lim\limits_{x\to c^+}f(x)$ uses values greater than $c$; and $\lim\limits_{x\to c}f(x)$ exists exactly when both one-sided limits exist and are equal. Neither limit depends on the value $f(c)$ itself — that value may even be missing. Continuity is the requirement that the value agrees with where the function is heading.

📘 Definition 5.1 — Continuity at a point

Let $f$ be a real function and let $c$ be a point in the domain of $f$. Then $f$ is continuous at $c$ if $$\lim_{x\to c}f(x)=f(c).$$ In detail, all three of the following must hold:

  1. $f(c)$ is defined (that is, $c$ is in the domain);
  2. $\lim\limits_{x\to c}f(x)$ exists, i.e. $\lim\limits_{x\to c^-}f(x)=\lim\limits_{x\to c^+}f(x)$ (both finite);
  3. this common value equals $f(c)$.

Briefly: LHL $=$ RHL $=f(c)$. If $c$ is in the domain and $f$ is not continuous at $c$, then $c$ is called a point of discontinuity of $f$. If $c$ is the left end-point $a$ of a domain $[a,b]$, only the right-hand limit is available, and $f$ is continuous at $a$ when $\lim\limits_{x\to a^+}f(x)=f(a)$; similarly at a right end-point $b$ we need $\lim\limits_{x\to b^-}f(x)=f(b)$.

When continuity fails, the way it fails is worth naming, because it tells you what (if anything) can be repaired:

removableLHL $=$ RHL, but the common value $\ne f(c)$ — a stray point. Redefining $f(c)$ repairs it.
jumpLHL and RHL are finite but different — the graph steps up or down. No choice of $f(c)$ helps.
infinite$f(x)$ grows without bound on at least one side — the graph shoots off along a vertical line.
✏️ Example 1 — Testing continuity at a point
Check whether each function is continuous at the given point.
(a) $f(x)=x^3-2x+1$ at $x=2$   (b) $f(x)=\dfrac{x^2-9}{x-3}$ for $x\ne3$, $f(3)=5$, at $x=3$
(c) $f(x)=\begin{cases}x+3,&x\le1\\ x^2+1,&x\gt1\end{cases}$ at $x=1$   (d) $f(x)=\begin{cases}x^2,&x\le2\\ 3x-2,&x\gt2\end{cases}$ at $x=2$
(a) $f(2)=8-4+1=5$. For a polynomial the limit is found by direct substitution (Class 11), so $\lim\limits_{x\to2}f(x)=5=f(2)$. Continuous at $2$.
(b) For $x\ne3$, $f(x)=\dfrac{(x-3)(x+3)}{x-3}=x+3$, so $\lim\limits_{x\to3}f(x)=6$. But $f(3)=5\ne6$. Not continuous at $3$ — a removable discontinuity: had we defined $f(3)=6$, $f$ would be continuous there.
(c) LHL $=\lim\limits_{x\to1^-}(x+3)=4$ and RHL $=\lim\limits_{x\to1^+}(x^2+1)=2$. They differ, so the limit does not exist. Not continuous at $1$ — a jump of size $2$. (Note $f(1)=4$, so $f$ agrees with its left-hand limit only.)
(d) LHL $=\lim\limits_{x\to2^-}x^2=4$, RHL $=\lim\limits_{x\to2^+}(3x-2)=4$, and $f(2)=2^2=4$. LHL $=$ RHL $=f(2)$: continuous at $2$.
Check Numerically for (c): $f(0.999)=3.999$ and $f(1.001)=2.002001$ — the two sides head for $4$ and $2$.
✏️ Example 2 — Modulus, signum and greatest integer
(a) Show that $f(x)=|x|$ is continuous at $x=0$. (b) Show that the signum function, $f(x)=\dfrac{|x|}{x}$ for $x\ne0$ and $f(0)=0$, is not continuous at $0$. (c) Find all points of discontinuity of the greatest integer function $f(x)=[x]$.
(a) For $x\lt0$, $|x|=-x$, so LHL $=\lim\limits_{x\to0^-}(-x)=0$; for $x\gt0$, $|x|=x$, so RHL $=0$; and $f(0)=0$. Continuous at $0$ — the graph has a corner there, but no break.
(b) For $x\lt0$, $\dfrac{|x|}{x}=\dfrac{-x}{x}=-1$; for $x\gt0$ it is $1$. So LHL $=-1\ne1=$ RHL: not continuous at $0$, whatever value $f(0)$ is given.
(c) integers Let $n$ be an integer. For $n-1\le x\lt n$ we have $[x]=n-1$, so LHL $=n-1$; for $n\le x\lt n+1$, $[x]=n$, so RHL $=n=f(n)$. LHL $\ne$ RHL, so $[x]$ is discontinuous at every integer.
(c) others If $c$ is not an integer, $[x]$ is the constant $[c]$ on a small interval around $c$ (for example $[x]=2$ for all $x$ near $2.7$), so $\lim\limits_{x\to c}[x]=[c]$: continuous. Points of discontinuity: exactly the integers.
Interactive 1 — The Continuity Checker

Pick a function and drag the orange point $c$ along the $x$-axis (or use the slider). The red tracer approaches $c$ from the left and the blue one from the right as $h$ shrinks; the readouts show where each side is heading and the value $f(c)$. Park $c$ at the special point of each function — and also somewhere else — and read the verdict.

$c$—
$h$—
$f(c-h)$ → LHL—
$f(c+h)$ → RHL—
$f(c)$—

Continuous Functions and Points of Discontinuity

A single point is rarely the whole story. We usually want to know where a function is continuous.

📘 Definition 5.2 — Continuous function; continuity on an interval

A real function $f$ is a continuous function if it is continuous at every point of its domain.

$f$ is continuous on an open interval $(a,b)$ if it is continuous at every point of $(a,b)$. It is continuous on a closed interval $[a,b]$ if it is continuous on $(a,b)$, and also $\lim\limits_{x\to a^+}f(x)=f(a)$ and $\lim\limits_{x\to b^-}f(x)=f(b)$ (continuity from the right at $a$ and from the left at $b$).

Notice the words “of its domain”. A point where $f$ is not defined is not a point of discontinuity: there is simply no value there to test. So $f(x)=\dfrac1x$, whose graph is in two separate pieces, is nevertheless a continuous function — it is continuous at every $x\ne0$, and $0$ is not in its domain. The “draw it without lifting the pen” picture is reliable only when the domain is a single interval.

🧭 Problem-solving strategy — locating points of discontinuity
  1. Write down the domain. Points outside it are never points of discontinuity.
  2. On each open interval where $f$ is given by a single polynomial, trigonometric or other standard formula, $f$ is continuous (Theorems 5.1–5.3 below make this official).
  3. Test the remaining “joining points” one at a time: compute LHL, RHL and $f(c)$.
  4. For $[\ \cdot\ ]$, test where the inside becomes an integer; for $|\ \cdot\ |$, the modulus itself never causes a break.
✏️ Example 3 — Find every point of discontinuity
(a) $f(x)=\begin{cases}2x+3,&x\le2\\ 2x-3,&x\gt2\end{cases}$   (b) $f(x)=\dfrac1x$   (c) $f(x)=\begin{cases}x^3-3,&x\le2\\ x^2+1,&x\gt2\end{cases}$   (d) $f(x)=\begin{cases}\dfrac{x}{|x|},&x\lt0\\ -1,&x\ge0\end{cases}$
(a) Domain $\mathbb R$. For $x\lt2$ and for $x\gt2$, $f$ is a polynomial, hence continuous. At $x=2$: LHL $=7=f(2)$, RHL $=\lim\limits_{x\to2^+}(2x-3)=1$. Only point of discontinuity: $x=2$.
(b) Domain $\mathbb R-\{0\}$. For $c\ne0$, $\lim\limits_{x\to c}\dfrac1x=\dfrac1c=f(c)$. So $f$ is continuous at every point of its domain: a continuous function with no point of discontinuity.
(c) Both pieces are polynomials. At $x=2$: LHL $=8-3=5=f(2)$ and RHL $=4+1=5$. Continuous at $2$ as well, so $f$ is continuous everywhere — the formula changes at $2$ but the graph does not break.
(d) For $x\lt0$, $\dfrac{x}{|x|}=\dfrac{x}{-x}=-1$. So $f(x)=-1$ for every real $x$: a constant function, continuous everywhere. Always simplify the pieces before testing.
✏️ Example 4 — Choosing constants to make a function continuous
(a) Find $k$ so that $f(x)=\begin{cases}kx^2,&x\le2\\ 3,&x\gt2\end{cases}$ is continuous at $x=2$.
(b) Find $a$ and $b$ so that $f(x)=\begin{cases}4,&x\le1\\ ax+b,&1\lt x\lt5\\ 16,&x\ge5\end{cases}$ is a continuous function.
(c) Find $k$ so that $f(x)=\begin{cases}kx+1,&x\le\pi\\ \cos x,&x\gt\pi\end{cases}$ is continuous at $x=\pi$.
(a) LHL $=f(2)=4k$ and RHL $=3$. Continuity needs $4k=3$, so $k=\dfrac34$.
(b) at 1 LHL $=f(1)=4$, RHL $=a+b$. So $a+b=4$.
(b) at 5 LHL $=5a+b$, RHL $=f(5)=16$. So $5a+b=16$. Subtracting, $4a=12$: $a=3$, $b=1$. Elsewhere each piece is a polynomial, so with these values $f$ is continuous everywhere.
(c) LHL $=f(\pi)=k\pi+1$ and RHL $=\cos\pi=-1$. So $k\pi+1=-1$ and $k=-\dfrac{2}{\pi}$.
Check (b): the middle piece $3x+1$ gives $4$ at $x=1$ and $16$ at $x=5$ ✓. Interactive 2 lets you hunt for such constants by sliding them.
Interactive 2 — Close the Gap: Tune the Constant

Each function contains a constant $k$. Move the slider (or press Play to sweep it) and watch the left piece and the right piece at the joining point. The gap readout is RHL − LHL (or limit − value). Only one value of $k$ closes the gap — find it, then check it by solving the equation “LHL = RHL = $f(c)$”.

$k$—
LHL—
RHL—
$f(c)$—
gap—

The Algebra of Continuous Functions and Composition

Testing every point with limits would be exhausting. Fortunately continuity survives the usual ways of combining functions, so a handful of basic facts covers almost everything.

📐 Theorem 5.1 — Algebra of continuous functions

Let $f$ and $g$ be real functions that are both continuous at a point $c$ (so $c$ is in both domains). Then

  1. $f+g$ and $f-g$ are continuous at $c$;
  2. $f\cdot g$ is continuous at $c$, and so is $\lambda f$ for every real constant $\lambda$;
  3. $\dfrac fg$ is continuous at $c$, provided $g(c)\ne0$.

Proof (sum). By the algebra of limits (Class 11) and the continuity of $f$ and $g$, $$\lim_{x\to c}(f+g)(x)=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)=f(c)+g(c)=(f+g)(c).$$ The other parts are proved the same way with the product and quotient rules for limits; the quotient rule needs $\lim\limits_{x\to c}g(x)=g(c)\ne0$. $\blacksquare$

📐 Theorem 5.2 — Standard continuous functions

(i) Every constant function, the identity $x$, and every polynomial $a_0+a_1x+\dots+a_nx^n$ is continuous at every real number. (ii) A rational function $\dfrac{p(x)}{q(x)}$ ($p,q$ polynomials) is continuous at every point where $q(x)\ne0$, i.e. it is a continuous function. (iii) $\sin x$ and $\cos x$ are continuous at every real number; $\tan x$, $\cot x$, $\sec x$, $\operatorname{cosec}x$ are continuous at every point of their domains. (iv) $|x|$ is continuous at every real number.

Why. (i) Constants and $x$ are continuous (their limits are found by substitution); a polynomial is built from them by sums and products, so Theorem 5.1 applies. (ii) follows from (i) and the quotient part of Theorem 5.1. (iii) For any $c$, put $x=c+h$: $$\lim_{h\to0}\sin(c+h)=\lim_{h\to0}(\sin c\cos h+\cos c\sin h)=\sin c\cdot1+\cos c\cdot0=\sin c,$$ using $\cos h\to1$ and $\sin h\to0$; $\cos x$ is similar, and $\tan x=\dfrac{\sin x}{\cos x}$ etc. are quotients with non-zero denominators on their domains. (iv) For $c\gt0$, $|x|=x$ near $c$; for $c\lt0$, $|x|=-x$ near $c$; at $0$ see Example 2(a). $\blacksquare$

Many functions are built in layers: in $\sin(x^2)$ we first square, then take the sine. If $g$ and $f$ are real functions, the function $x\mapsto f(g(x))$ — “first $g$, then $f$” — is called the composite of $f$ and $g$ and is written $f\circ g$; it is defined at those $x$ for which $g(x)$ lies in the domain of $f$.

📐 Theorem 5.3 — Continuity of a composite function

Let $f$ and $g$ be real functions such that $f\circ g$ is defined at $c$. If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then $f\circ g$ is continuous at $c$: $$\lim_{x\to c}f(g(x))=f\Big(\lim_{x\to c}g(x)\Big)=f(g(c)).$$

Why. As $x\to c$, continuity of $g$ makes $g(x)\to g(c)$; and as the input of $f$ approaches $g(c)$, continuity of $f$ at $g(c)$ makes $f(g(x))\to f(g(c))$. $\blacksquare$

✏️ Example 5 — Continuity without computing limits
Show that each function is continuous: (a) $\sin(x^2)$   (b) $|1-x+|x||$   (c) $|x|-|x+1|$   (d) $\dfrac{x^2+1}{x^2-5x+6}$. In (d), is $x=2$ a point of discontinuity?
(a) $\sin(x^2)=f(g(x))$ with $g(x)=x^2$ (a polynomial) and $f(u)=\sin u$, both continuous everywhere. By Theorem 5.3, $\sin(x^2)$ is continuous at every real $x$.
(b) $u(x)=1-x+|x|$ is a sum of continuous functions (Theorems 5.1, 5.2). Then $|u(x)|$ is the composite of $u$ with the continuous function $|\ \cdot\ |$, so it is continuous everywhere.
(c) $|x|$ is continuous; $|x+1|$ is the composite of $x+1$ and $|\ \cdot\ |$, so it is continuous; their difference is continuous (Theorem 5.1). Continuous everywhere — even though its graph has two corners, at $-1$ and $0$.
(d) A rational function, with denominator $(x-2)(x-3)$. Its domain is $\mathbb R-\{2,3\}$ and it is continuous at every point of that domain (Theorem 5.2(ii)) — a continuous function. $x=2$ is not a point of discontinuity, because $f(2)$ is not defined; the question “is $f$ continuous at $2$?” does not arise.
✏️ Example 6 — Constants from standard limits
Find $k$ so that each function is continuous at $x=0$.
(a) $f(x)=\dfrac{\sin3x}{x}$ for $x\ne0$, $f(0)=k$   (b) $f(x)=\dfrac{1-\cos4x}{8x^2}$ for $x\ne0$, $f(0)=k$
(a) $\lim\limits_{x\to0}\dfrac{\sin3x}{x}=\lim\limits_{x\to0}3\cdot\dfrac{\sin3x}{3x}=3\cdot1=3$, using $\lim\limits_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$ with $\theta=3x$. Continuity needs $k=3$.
(b) $1-\cos4x=2\sin^22x$, so $\dfrac{1-\cos4x}{8x^2}=\dfrac{2\sin^22x}{8x^2}=\dfrac{2\cdot4}{8}\left(\dfrac{\sin2x}{2x}\right)^2\to1\cdot1^2=1$. So $k=1$.
Check At $x=0.001$: $\dfrac{\sin0.003}{0.001}\approx2.999996$ and $\dfrac{1-\cos0.004}{8\times10^{-6}}\approx0.999999$ ✓.
⚠️ Common mistakes when studying alone

1. Calling a point outside the domain a discontinuity. $\tan x$ is a continuous function; $x=\tfrac\pi2$ is not a point of discontinuity, because $\tan\tfrac\pi2$ does not exist. Board answers are marked on exactly this point.

2. Checking only one side. For a piecewise function, LHL, RHL and $f(c)$ are needed, and you must use the correct piece for each: the piece with “$x\le c$” gives both the LHL and $f(c)$.

3. “The limit exists, so $f$ is continuous.” Not unless the limit equals $f(c)$ — Example 1(b) has a limit and is still discontinuous.

4. Simplifying across the point. $\dfrac{x^2-9}{x-3}=x+3$ only for $x\ne3$. The simplification gives the limit; the value at $3$ must be read from the definition of $f$.

5. Confusing a corner with a break. $|x|$ has a sharp corner at $0$ but is continuous there. Corners matter for differentiability, the subject of the next section.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026; not examined in the CBSE board exam. JEE loves functions built from $[x]$, where a jump of one factor can be cancelled by a zero of another. The method is always the same: at each integer $n$ use $[x]=n-1$ just to the left and $[x]=n$ at and just to the right.

Worked example. Where are (a) $f(x)=[x]\sin\pi x$ and (b) $g(x)=[x]+[-x]$ continuous?

(a) At a non-integer, $f$ is a constant times $\sin\pi x$ nearby: continuous. At an integer $n$: LHL $=(n-1)\sin n\pi=0$, RHL $=n\sin n\pi=0$, $f(n)=0$. So $f$ is continuous everywhere — the zeros of $\sin\pi x$ sit exactly on the jumps of $[x]$.

(b) If $x$ is an integer, $[x]+[-x]=x-x=0$. If not, $[-x]=-[x]-1$, so $g(x)=-1$. At every integer the limit is $-1$ but the value is $0$: a removable discontinuity at every integer, continuous elsewhere.

Exercises

I
Is the function continuous at the given point? Give LHL, RHL and the value where relevant.
  • 1 $f(x)=2x^2-3x+5$ at $x=1$
  • 2 $f(x)=\dfrac{x^2-25}{x-5}$ for $x\ne5$, $f(5)=10$, at $x=5$
  • 3 $f(x)=\dfrac{x^2-25}{x-5}$ for $x\ne5$, $f(5)=5$, at $x=5$
  • 4 $f(x)=\begin{cases}3x-2,&x\le0\\ x+1,&x\gt0\end{cases}$ at $x=0$
  • 5 $f(x)=\begin{cases}x^2,&x\le1\\ 2x-1,&x\gt1\end{cases}$ at $x=1$
  • 6 $f(x)=|x-3|$ at $x=3$
  • 7 $f(x)=x-[x]$ at $x=2$
  • 8 $f(x)=\dfrac{\sin x}{x}$ for $x\ne0$, $f(0)=1$, at $x=0$
  • 9 Is $x=2$ a point of discontinuity of $f(x)=\dfrac{1}{x-2}$? Explain.
1 Polynomial: $\lim\limits_{x\to1}f(x)=f(1)=2-3+5=4$. Continuous.
2 For $x\ne5$, $f(x)=x+5$, so the limit is $10=f(5)$. Continuous.
3 The limit is still $10$, but $f(5)=5\ne10$. Not continuous (removable).
4 LHL $=f(0)=-2$, RHL $=1$. Not continuous (jump of $3$).
5 LHL $=f(1)=1$, RHL $=2-1=1$. Continuous.
6 $|x-3|$ is the composite of $x-3$ and $|\ \cdot\ |$: continuous. Directly: LHL $=\lim(3-x)=0$, RHL $=\lim(x-3)=0$, $f(3)=0$.
7 For $1\le x\lt2$, $f(x)=x-1$, so LHL $=1$; for $2\le x\lt3$, $f(x)=x-2$, so RHL $=0=f(2)$. Not continuous.
8 $\lim\limits_{x\to0}\dfrac{\sin x}{x}=1=f(0)$. Continuous.
9 No. $2$ is not in the domain of $f$, so it cannot be a point of discontinuity; $f$ is continuous at every point of its domain $\mathbb R-\{2\}$.
II
Find all points of discontinuity (write “none” if the function is continuous).
  • 1 $f(x)=x^3-x^2+4$
  • 2 $f(x)=\begin{cases}x+1,&x\ge1\\ x^2+1,&x\lt1\end{cases}$
  • 3 $f(x)=\dfrac{|x|}{x}$ for $x\ne0$, $f(0)=1$
  • 4 $f(x)=\begin{cases}|x|+3,&x\le-3\\ -2x,&-3\lt x\lt3\\ 6x+2,&x\ge3\end{cases}$
  • 5 $f(x)=[x]$ on the interval $-2\lt x\lt2$
  • 6 $f(x)=\tan x$
  • 7 $f(x)=x^2\sin\dfrac1x$ for $x\ne0$, $f(0)=0$  (Hint: $-x^2\le x^2\sin\tfrac1x\le x^2$.)
  • 8 $f(x)=\begin{cases}x+2,&x\le1\\ x-2,&x\gt1\end{cases}$
1 A polynomial: none.
2 Away from $1$ both pieces are polynomials. At $1$: LHL $=1+1=2$, RHL $=f(1)=2$. None.
3 LHL $=-1$, RHL $=1$: discontinuous at $x=0$ only.
4 At $-3$: LHL $=f(-3)=3+3=6$, RHL $=-2(-3)=6$: continuous. At $3$: LHL $=-6$, RHL $=f(3)=20$: discontinuous. Only $x=3$.
5 The integers in $(-2,2)$: $x=-1,0,1$.
6 None: $\tan x=\dfrac{\sin x}{\cos x}$ is continuous wherever $\cos x\ne0$, which is its whole domain.
7 For $x\ne0$, $f$ is a product and composite of continuous functions. At $0$: $-x^2\le f(x)\le x^2$ and both bounds $\to0$, so by the sandwich theorem $\lim\limits_{x\to0}f(x)=0=f(0)$. None.
8 At $1$: LHL $=f(1)=3$, RHL $=-1$. Only $x=1$.
III
Find the constants that make each function continuous at the stated point (or everywhere).
  • 1 $f(x)=\begin{cases}kx+1,&x\le5\\ 3x-5,&x\gt5\end{cases}$ at $x=5$
  • 2 $f(x)=\begin{cases}k(x^2+2),&x\le0\\ 3x+1,&x\gt0\end{cases}$ at $x=0$
  • 3 $f(x)=\dfrac{x^2-16}{x-4}$ for $x\ne4$, $f(4)=k$, at $x=4$
  • 4 $f(x)=\dfrac{k\cos x}{\pi-2x}$ for $x\ne\tfrac\pi2$, $f\!\left(\tfrac\pi2\right)=3$, at $x=\tfrac\pi2$
  • 5 $f(x)=\dfrac{1-\cos kx}{x\sin x}$ for $x\ne0$, $f(0)=\tfrac12$, at $x=0$
  • 6 $f(x)=\begin{cases}7,&x\le1\\ ax+b,&1\lt x\lt4\\ 1,&x\ge4\end{cases}$, everywhere
  • 7 $f(x)=\begin{cases}-2\sin x,&x\le-\tfrac\pi2\\ a\sin x+b,&-\tfrac\pi2\lt x\lt\tfrac\pi2\\ \cos x,&x\ge\tfrac\pi2\end{cases}$, everywhere
1 LHL $=f(5)=5k+1$, RHL $=10$: $k=\dfrac95$.
2 LHL $=f(0)=2k$, RHL $=1$: $k=\dfrac12$.
3 For $x\ne4$, $f(x)=x+4\to8$: $k=8$.
4 Put $x=\tfrac\pi2+h$: $\cos x=-\sin h$ and $\pi-2x=-2h$, so $f=\dfrac{k\sin h}{2h}\to\dfrac k2$. Then $\dfrac k2=3$: $k=6$.
5 $1-\cos kx=2\sin^2\tfrac{kx}{2}$, so $f(x)=\dfrac{2\sin^2(kx/2)}{x\sin x}=\dfrac{k^2}{2}\cdot\left(\dfrac{\sin(kx/2)}{kx/2}\right)^2\cdot\dfrac{x}{\sin x}\to\dfrac{k^2}{2}$ (for $k\ne0$). So $\dfrac{k^2}{2}=\dfrac12$: $k=\pm1$. ($k=0$ gives $f=0$ near $0$, which fails.)
6 At $1$: $a+b=7$. At $4$: $4a+b=1$. So $3a=-6$: $a=-2$, $b=9$.
7 At $-\tfrac\pi2$: LHL $=f(-\tfrac\pi2)=-2\sin(-\tfrac\pi2)=2$ and RHL $=-a+b$, so $b-a=2$. At $\tfrac\pi2$: LHL $=a+b$, RHL $=f(\tfrac\pi2)=\cos\tfrac\pi2=0$, so $a+b=0$. Hence $a=-1$, $b=1$.
IV
Combining functions, reasoning and modelling.
  • 1 Show that $f(x)=\cos|x|$ is continuous everywhere. (Can you write it more simply?)
  • 2 Show that $f(x)=\sin|x|$ is continuous everywhere.
  • 3 Show that $f(x)=|x|+|x-1|$ is continuous everywhere, and write it without modulus signs on each of $x\lt0$, $0\le x\lt1$, $x\ge1$.
  • 4 Give two functions, each discontinuous at $0$, whose sum is continuous at $0$.
  • 5 $f$ is continuous at $c$ and $g$ is not. Can $f+g$ be continuous at $c$? Justify.
  • 6 Parking. A mall charges ₹20 for the first hour or part of it, and ₹15 for each further hour or part. Let $P(t)$ be the fee for a stay of $t$ hours, $0\lt t\le5$. Find $P(1)$ and $P(1.25)$, and list the points of discontinuity of $P$.
  • 7 Water tank. The depth of water in a tank $t$ minutes after a pump starts is $h(t)=0.04t$ m for $0\le t\le30$ and $h(t)=a-0.02t$ m for $30\lt t\le90$ (the tank is then drained). Find $a$ so that $h$ is continuous, and the depth at $t=90$.
1 $\cos|x|$ is the composite of the continuous functions $|x|$ and $\cos$ (Theorem 5.3). In fact $\cos|x|=\cos x$, since cosine is even.
2 The composite of $|x|$ and $\sin$, both continuous: continuous everywhere. (Unlike $\cos|x|$, it is not equal to $\sin x$ for $x\lt0$.)
3 A sum of continuous functions. $f(x)=-2x+1$ for $x\lt0$; $f(x)=1$ for $0\le x\lt1$; $f(x)=2x-1$ for $x\ge1$. The pieces agree at $0$ (value $1$) and at $1$ (value $1$).
4 For example $f(x)=\dfrac{|x|}{x}$ ($x\ne0$), $f(0)=0$, and $g=-f$. Then $f+g=0$, a constant, which is continuous.
5 No. If $f+g$ were continuous at $c$, then $g=(f+g)-f$ would be a difference of functions continuous at $c$, hence continuous at $c$ by Theorem 5.1 — a contradiction.
6 $P(1)=$ ₹20 (one hour is “the first hour”) and $P(1.25)=$ ₹35. The fee is constant on $(0,1]$, $(1,2]$, $(2,3]$, $(3,4]$, $(4,5]$ and jumps by ₹15 just after each whole hour, so the points of discontinuity are $t=1,2,3,4$. (At each, $P$ equals the LHL but not the RHL.)
7 $h(30)=1.2$ m from the first formula; the second gives RHL $=a-0.6$. So $a=1.8$, and $h(90)=1.8-1.8=0$ m: the tank is empty after $90$ minutes.
🎯 Exam corner Board · 1 markBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: the Board almost always sets a 1-mark “find $k$” question and a 3-mark problem with two unknown constants at two joining points; a case study may model a tariff or a fee. JEE tests $[x]$- and modulus-type functions and counts points of discontinuity.

E1. Board · 1 mark The value of $k$ for which $f(x)=\dfrac{\sin5x}{3x}$ ($x\ne0$), $f(0)=k$, is continuous at $x=0$ is  (A) $\tfrac35$  (B) $\tfrac53$  (C) $5$  (D) $15$

Idea $\dfrac{\sin5x}{3x}=\dfrac53\cdot\dfrac{\sin5x}{5x}\to\dfrac53$.
Answer (B). Trap (A) inverts the ratio; (C) forgets the $3$ in the denominator.

E2. Board · 1 mark Assertion–Reason. Assertion (A): $f(x)=\dfrac{1}{x-3}$ is a continuous function. Reason (R): a real function is continuous if it is continuous at every point of its domain.  (A) Both A and R are true and R is the correct explanation of A  (B) Both A and R are true but R is not the correct explanation of A  (C) A is true but R is false  (D) A is false but R is true

Idea R is Definition 5.2. The domain of $f$ is $\mathbb R-\{3\}$, and $f$ is a rational function, continuous at every point of it — so A is true because of R.
Answer (A). The tempting (D) treats $x=3$ as a point of discontinuity, but $3$ is not in the domain.

E3. Board · 3 marks Find $a$ and $b$ if $f(x)=\begin{cases}5ax-2b,&x\lt1\\ 11,&x=1\\ 3ax+b,&x\gt1\end{cases}$ is continuous at $x=1$.

Limits LHL $=\lim\limits_{x\to1^-}(5ax-2b)=5a-2b$; RHL $=\lim\limits_{x\to1^+}(3ax+b)=3a+b$; $f(1)=11$. (1 mark)
Equations $3a+b=11$ and $5a-2b=11$. From the first, $b=11-3a$; then $5a-22+6a=11$, so $11a=33$. (1 mark)
Answer $a=3$, $b=2$. Check: $5(3)-2(2)=11$ ✓ and $3(3)+2=11$ ✓. (1 mark)

E4. Board · Case study A state electricity board charges ₹3 per unit for the first $100$ units used in a month and ₹5 per unit for every unit beyond $100$ (a “telescopic” tariff), so the bill for $u$ units is $B(u)=3u$ for $0\le u\le100$ and $B(u)=300+5(u-100)$ for $u\gt100$. (i) Show that $B$ is continuous at $u=100$. [2 marks] (ii) A rival board charges $5u$ for the whole consumption once $u\gt100$ (and $3u$ for $u\le100$). What happens to its bill at $u=100$? [1 mark] (iii) The rival instead charges $5u-c$ for $u\gt100$. Find the rebate $c$ that makes its bill continuous. [1 mark]

(i) LHL $=\lim\limits_{u\to100^-}3u=300=B(100)$, RHL $=\lim\limits_{u\to100^+}[300+5(u-100)]=300$. LHL $=$ RHL $=B(100)$: continuous.
(ii) LHL $=300$ but RHL $=500$: the bill jumps by ₹200 — using one extra unit beyond $100$ costs ₹205. Discontinuous at $u=100$.
(iii) Need $500-c=300$, so $c=200$, i.e. a rebate of ₹200. (Then $5u-200=300+5(u-100)$: it is the telescopic tariff again.)

E5. JEE Main The number of points in the open interval $(0,3)$ at which $f(x)=[x^2]$ is discontinuous is  (A) $2$  (B) $3$  (C) $8$  (D) $9$

Method On $(0,3)$, $x^2$ is continuous and strictly increasing with values in $(0,9)$. $[x^2]$ jumps exactly where $x^2$ passes an integer $n=1,2,\dots,8$, i.e. at $x=\sqrt n$: $1,\sqrt2,\sqrt3,2,\sqrt5,\sqrt6,\sqrt7,2\sqrt2$ — all inside $(0,3)$. At each, LHL $=n-1$ and RHL $=n$.
Answer (C) $8$. Trap (A): counting only the integers $x=1,2$; (D) wrongly includes $x=3$, which is not in the open interval.
⏱ Fast route Count the integers strictly between $0^2$ and $3^2$: $1,\dots,8$.

E6. JEE Advanced (One or more options correct.) Let $f(x)=[x]\sin\pi x$, $g(x)=[x]+[-x]$ and $h(x)=x[x]$. Then  (A) $f$ is continuous on $\mathbb R$  (B) $g$ is continuous at $x=1$  (C) $h$ is continuous at $x=0$  (D) $g$ has a removable discontinuity at every integer

(A) True At an integer $n$ both one-sided limits are $(n-1)\cdot0$ and $n\cdot0$, i.e. $0=f(n)$ (see the 🚀 box).
(B) False, (D) True $g(x)=-1$ for non-integers and $g(n)=0$, so $\lim\limits_{x\to n}g(x)=-1\ne g(n)$: a removable discontinuity at every integer, including $1$.
(C) True For $-1\le x\lt0$, $h(x)=-x\to0$; for $0\le x\lt1$, $h(x)=0$; $h(0)=0$. LHL $=$ RHL $=h(0)$. (At $x=1$, by contrast, LHL $=\lim\limits_{x\to1^-}x\cdot0=0$ but RHL $=\lim\limits_{x\to1^+}x\cdot1=1$: $h$ jumps there.)
Answer (A), (C), (D).
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Is your household electricity bill a continuous function of the number of units used — and if not, where does it jump, and by how much?

Why it matters. Slab tariffs are used for electricity, water and income tax all over the world. A tariff whose bill jumps at a slab boundary can make one extra unit cost hundreds of rupees, which feels unfair and encourages people to under-report. Regulators therefore design “telescopic” slabs, which keep the bill continuous even though the price per unit jumps.

What to do.

  1. Find the domestic tariff of your state's electricity distribution company (it is published on the company's or the state regulatory commission's website) or use a recent bill at home. Note the slabs, the price per unit in each, and any fixed charge that depends on the slab.
  2. Write the bill $B(u)$ as a piecewise function of the monthly units $u$, and test continuity at every slab boundary using LHL, RHL and $B(u)$.
  3. In a spreadsheet, tabulate $B(u)$ for $u=0,1,\dots,500$ and plot it. Mark every jump and its size.
  4. Compute the “marginal cost” $B(u+1)-B(u)$ and plot it as well. Which graph has jumps for a telescopic tariff, and which does not?

Tools: Spreadsheet; your state's published tariff or a household bill.

Hand in: The piecewise formula, the continuity check at each boundary, the two plots, and a short paragraph: is the tariff fair at the boundaries? Suggest a change of fixed charges that would make the bill continuous.

Going further. Income tax in the new Indian regime is also slab-based. Is tax owed a continuous function of income? Look up “marginal relief” and explain, in terms of continuity, the problem it solves.

Instructor note: a purely telescopic energy charge always gives a continuous bill (each piece agrees at the boundary), while the marginal price jumps. Many real tariffs add slab-dependent fixed charges, which create genuine jump discontinuities; E4 in the Exam corner is a ready example ($c=200$ restores continuity). Income-tax rebates that cut off sharply create jumps, and marginal relief is exactly the patch that makes tax continuous at the cut-off.

🧪 Self-check quiz

0/5 correct

Q1. A function $f$ is continuous at a point $c$ of its domain exactly when

Q2. The points of discontinuity of $f(x)=\dfrac{1}{x-2}$ are

Q3. The greatest integer function $[x]$ is discontinuous at

Q4. $f(x)=\dfrac{x^2-1}{x-1}$ for $x\ne1$ and $f(1)=k$. $f$ is continuous at $1$ when $k=$

Q5. $f$ and $g$ are continuous at $c$. Which of these might fail to be continuous at $c$?

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