NCERT Class 12 Mathematics — Continuity and Differentiability, Exercise 5.1. All 34 questions solved.

A function $f$ is continuous at a point $c$ when three things all hold and agree:

$$\lim_{x \to c^-}f(x) = \lim_{x \to c^+}f(x) = f(c)$$

That is: the left-hand limit exists, the right-hand limit exists, they are equal, and their common value is the function’s actual value there.

Key insight. For a piecewise function, only the joining points can cause trouble. Everywhere else the function agrees locally with a polynomial, trigonometric or exponential expression, all of which are continuous. So the whole exercise reduces to checking one point per join.

At a join, compute the two one-sided limits by substituting into the branch that applies on that side — and then compare both with $f(c)$, taking care to read which branch the $\le$ or $\ge$ sign puts $c$ into. Question 15’s join at $x = 1$ is decided entirely by the fact that $f(1) = 0$ from the middle branch while the right-hand limit is $4$.

Question 1

Prove that the function $f(x) = 5x – 3$ is continuous at $x = 0$, at $x = -3$ and at $x = 5$.

Solution. For any real $c$,

$$\lim_{x \to c}(5x – 3) = 5c – 3 = f(c)$$

since a polynomial’s limit is found by direct substitution. Taking $c = 0$, $-3$ and $5$ gives $f(0) = -3$, $f(-3) = -18$ and $f(5) = 22$, each equal to the corresponding limit.

Question 2

Examine the continuity of the function $f(x) = 2x^2 – 1$ at $x = 3$.

Solution. $\displaystyle\lim_{x\to3}\left(2x^2-1\right) = 2(9) – 1 = 17$, and $f(3) = 17$.

Question 3

Examine the following functions for continuity.

(a) $f(x) = x – 5$   (b) $f(x) = \dfrac{1}{x-5}$, $x \ne 5$   (c) $f(x) = \dfrac{x^2-25}{x+5}$, $x \ne -5$   (d) $f(x) = |x-5|$

Solution.

(a) A polynomial, continuous everywhere.

(b) A rational function, continuous at every point of its domain. Note that $x = 5$ is not in the domain, so there is nothing to check there — a function cannot be discontinuous at a point where it is not defined.

(c) Again rational, continuous on its domain $\mathbf{R} – \{-5\}$.

(d) $|x-5|$ is the composition of the continuous functions $x \mapsto x-5$ and $x \mapsto |x|$, so it is continuous everywhere — including at $x = 5$, where it has a corner but no break.

Question 4

Prove that the function $f(x) = x^n$ is continuous at $x = n$, where $n$ is a positive integer.

Solution. For a positive integer $n$, $x^n$ is a polynomial, and

$$\lim_{x \to n}x^n = n^n = f(n)$$

by the algebra of limits (the limit of a product is the product of the limits, applied $n$ times). Hence $f$ is continuous at $x = n$.

Question 5

Is the function $f$ defined by $f(x) = \begin{cases}x, & x \le 1 \\ 5, & x > 1\end{cases}$ continuous at $x = 0$? At $x = 1$? At $x = 2$?

Solution. Only $x = 1$ is a joining point.

At $x = 0$: both the limit and the value come from the branch $f(x) = x$, so $f$ is continuous.

At $x = 1$: $\displaystyle\lim_{x\to1^-}f(x) = 1$ but $\displaystyle\lim_{x\to1^+}f(x) = 5$. The one-sided limits disagree, so $f$ is not continuous.

At $x = 2$: locally $f$ is the constant $5$, hence continuous.

Find all points of discontinuity of $f$, where $f$ is defined by:

Question 6

$f(x) = \begin{cases}2x + 3, & x \le 2 \\ 2x – 3, & x > 2\end{cases}$

Solution. At $x = 2$:

$$\lim_{x\to2^-}f(x) = 7, \qquad \lim_{x\to2^+}f(x) = 1$$

The two differ by the constant $6$, which is the gap between the branches.

Question 7

$f(x) = \begin{cases}|x| + 3, & x \le -3 \\ -2x, & -3 < x < 3 \\ 6x + 2, & x \ge 3\end{cases}$

Solution. Two joins to check.

At $x = -3$: from the left, $|x| + 3 \to 3 + 3 = 6$; from the right, $-2x \to 6$; and $f(-3) = 6$. Continuous.

At $x = 3$: from the left, $-2x \to -6$; from the right, $6x + 2 \to 20$. Discontinuous.

Question 8

$f(x) = \begin{cases}\dfrac{|x|}{x}, & x \ne 0 \\[4pt] 0, & x = 0\end{cases}$

Solution. For $x > 0$, $\tfrac{|x|}{x} = 1$; for $x < 0$ it is $-1$.

$$\lim_{x\to0^-}f(x) = -1, \qquad \lim_{x\to0^+}f(x) = 1$$

The value $f(0) = 0$ is a third number again, but the disagreement of the limits already settles it.

Question 9

$f(x) = \begin{cases}\dfrac{x}{|x|}, & x < 0 \\[4pt] -1, & x \ge 0\end{cases}$

Solution. For $x < 0$ we have $|x| = -x$, so $\tfrac{x}{|x|} = -1$. Both branches therefore give $-1$, and $f$ is the constant function $-1$ in disguise.

This is the one place in the exercise where reading the definition carefully saves all the work.

Question 10

$f(x) = \begin{cases}x + 1, & x \ge 1 \\ x^2 + 1, & x < 1\end{cases}$

Solution. At $x = 1$: the left-hand limit is $1 + 1 = 2$, the right-hand limit is $1 + 1 = 2$, and $f(1) = 2$.

Question 11

$f(x) = \begin{cases}x^3 – 3, & x \le 2 \\ x^2 + 1, & x > 2\end{cases}$

Solution. At $x = 2$: from the left, $8 – 3 = 5$; from the right, $4 + 1 = 5$; and $f(2) = 5$.

Question 12

$f(x) = \begin{cases}x^{10} – 1, & x \le 1 \\ x^2, & x > 1\end{cases}$

Solution. At $x = 1$: the left-hand limit is $1 – 1 = 0$, but the right-hand limit is $1$.

Question 13

Is the function defined by $f(x) = \begin{cases}x + 5, & x \le 1 \\ x – 5, & x > 1\end{cases}$ a continuous function?

Solution. At $x = 1$: the left-hand limit is $6$ and the right-hand limit is $-4$.

Discuss the continuity of the function $f$, where $f$ is defined by:

Question 14

$f(x) = \begin{cases}3, & 0 \le x \le 1 \\ 4, & 1 < x < 3 \\ 5, & 3 \le x \le 10\end{cases}$

Solution. A step function with two jumps.

At $x = 1$: left limit $3$, right limit $4$ — discontinuous.

At $x = 3$: left limit $4$, right limit $5$ — discontinuous.

Everywhere else on $[0, 10]$ the function is locally constant, hence continuous.

Question 15

$f(x) = \begin{cases}2x, & x < 0 \\ 0, & 0 \le x \le 1 \\ 4x, & x > 1\end{cases}$

Solution.

At $x = 0$: left limit $2(0) = 0$, right limit $0$, and $f(0) = 0$ — continuous.

At $x = 1$: $f(1) = 0$ from the middle branch, and the left limit is $0$; but the right limit is $4(1) = 4$ — discontinuous.

Question 16

$f(x) = \begin{cases}-2, & x \le -1 \\ 2x, & -1 < x \le 1 \\ 2, & x > 1\end{cases}$

Solution.

At $x = -1$: left limit $-2$, right limit $2(-1) = -2$, and $f(-1) = -2$.

At $x = 1$: left limit $2(1) = 2$, right limit $2$, and $f(1) = 2$.

Both joins match, because the middle branch was chosen to meet the outer constants exactly.

Question 17

Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x) = \begin{cases}ax + 1, & x \le 3 \\ bx + 3, & x > 3\end{cases}$ is continuous at $x = 3$.

Solution. Continuity requires the two branches to meet at $x = 3$:

$$3a + 1 = 3b + 3 \quad\Longrightarrow\quad 3a = 3b + 2$$

Question 18

For what value of $\lambda$ is the function defined by $f(x) = \begin{cases}\lambda\left(x^2 – 2x\right), & x \le 0 \\ 4x + 1, & x > 0\end{cases}$ continuous at $x = 0$? What about continuity at $x = 1$?

Solution. At $x = 0$:

$$\lim_{x\to0^-}f(x) = \lambda(0 – 0) = 0, \qquad \lim_{x\to0^+}f(x) = 1$$

The left-hand limit is $0$ whatever $\lambda$ is, because both terms of $x^2 – 2x$ vanish at $0$. So the limits can never agree.

At $x = 1$ the function is simply $4x + 1$ in a neighbourhood, a polynomial, so it is continuous there for every $\lambda$.

Question 19

Show that the function defined by $g(x) = x – [x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.

Solution. Let $n$ be any integer. Just to the left of $n$, say $x = n – h$ with small $h > 0$, the greatest integer is $n-1$, so $g(x) = n – h – (n-1) = 1 – h$, giving

$$\lim_{x\to n^-}g(x) = 1$$

Just to the right, $[x] = n$, so $g(x) = h$ and

$$\lim_{x\to n^+}g(x) = 0$$

Since $1 \ne 0$, $g$ is discontinuous at every integer. (It is continuous everywhere else, where $[x]$ is locally constant.)

Question 20

Is the function defined by $f(x) = x^2 – \sin x + 5$ continuous at $x = \pi$?

Solution. Each of $x^2$, $\sin x$ and the constant $5$ is continuous everywhere, and sums and differences of continuous functions are continuous. So

$$\lim_{x\to\pi}f(x) = \pi^2 – 0 + 5 = f(\pi)$$

Question 21

Discuss the continuity of the following functions:

(a) $f(x) = \sin x + \cos x$   (b) $f(x) = \sin x – \cos x$   (c) $f(x) = \sin x \cdot \cos x$

Solution. Both $\sin x$ and $\cos x$ are continuous on the whole of $\mathbf{R}$. The algebra of continuous functions says that the sum, difference and product of continuous functions are continuous. All three follow at once.

Question 22

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution. Cosine is continuous everywhere. The other three are quotients, and a quotient is continuous wherever its denominator is non-zero.

• $\operatorname{cosec}x = \tfrac{1}{\sin x}$ fails where $\sin x = 0$, that is at $x = n\pi$.
• $\sec x = \tfrac{1}{\cos x}$ fails where $\cos x = 0$, that is at $x = (2n+1)\tfrac{\pi}{2}$.
• $\cot x = \tfrac{\cos x}{\sin x}$ fails where $\sin x = 0$, that is at $x = n\pi$.

At those points the functions are simply not defined, so they are continuous on their whole domains.

Question 23

Find all points of discontinuity of $f$, where $f(x) = \begin{cases}\dfrac{\sin x}{x}, & x < 0 \\[4pt] x + 1, & x \ge 0\end{cases}$

Solution. At $x = 0$, use the standard limit $\displaystyle\lim_{x\to0}\frac{\sin x}{x} = 1$:

$$\lim_{x\to0^-}f(x) = 1, \qquad \lim_{x\to0^+}f(x) = 1, \qquad f(0) = 1$$

Away from $0$ both branches are continuous on their own domains.

Question 24

Determine if $f$ defined by $f(x) = \begin{cases}x^2\sin\dfrac{1}{x}, & x \ne 0 \\[4pt] 0, & x = 0\end{cases}$ is a continuous function.

Solution. The only doubtful point is $x = 0$, where $\sin\tfrac1x$ oscillates without limit. But it is bounded: $\left|\sin\tfrac1x\right| \le 1$, so

$$\left|x^2\sin\frac1x\right| \le x^2$$

As $x \to 0$ the right side tends to $0$, so by the sandwich theorem the left side does too. Hence the limit is $0 = f(0)$.

The $x^2$ factor is doing all the work: it squashes the oscillation to nothing.

Question 25

Examine the continuity of $f$, where $f$ is defined by $f(x) = \begin{cases}\sin x – \cos x, & x \ne 0 \\ -1, & x = 0\end{cases}$

Solution. At $x = 0$:

$$\lim_{x\to0}(\sin x – \cos x) = 0 – 1 = -1 = f(0)$$

The special value at $0$ was chosen to match, so nothing breaks.

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29.

Question 26

$f(x) = \begin{cases}\dfrac{k\cos x}{\pi – 2x}, & x \ne \tfrac{\pi}{2} \\[4pt] 3, & x = \tfrac{\pi}{2}\end{cases}$ at $x = \dfrac{\pi}{2}$

Solution. The fraction is $\tfrac00$ at $x = \tfrac{\pi}{2}$, so substitute $x = \tfrac{\pi}{2} + h$ and let $h \to 0$. Then $\cos x = -\sin h$ and $\pi – 2x = -2h$:

$$\lim_{h\to0}\frac{k(-\sin h)}{-2h} = \frac{k}{2}\lim_{h\to0}\frac{\sin h}{h} = \frac{k}{2}$$

Setting this equal to $f\left(\tfrac{\pi}{2}\right) = 3$ gives $k = 6$.

Question 27

$f(x) = \begin{cases}kx^2, & x \le 2 \\ 3, & x > 2\end{cases}$ at $x = 2$

Solution. $4k = 3$.

Question 28

$f(x) = \begin{cases}kx + 1, & x \le \pi \\ \cos x, & x > \pi\end{cases}$ at $x = \pi$

Solution. $k\pi + 1 = \cos\pi = -1$, so $k\pi = -2$.

Question 29

$f(x) = \begin{cases}kx + 1, & x \le 5 \\ 3x – 5, & x > 5\end{cases}$ at $x = 5$

Solution. $5k + 1 = 15 – 5 = 10$, so $5k = 9$.

Question 30

Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases}5, & x \le 2 \\ ax + b, & 2 < x < 10 \\ 21, & x \ge 10\end{cases}$ is a continuous function.

Solution. Two joins give two equations:

$$\text{at } x = 2: \quad 2a + b = 5$$

$$\text{at } x = 10: \quad 10a + b = 21$$

Subtracting, $8a = 16$, so $a = 2$ and then $b = 1$.

The middle branch is the straight line through $(2, 5)$ and $(10, 21)$ — which is exactly what “continuous” demands of it.

Question 31

Show that the function defined by $f(x) = \cos\left(x^2\right)$ is a continuous function.

Solution. Write $f = g \circ h$ with $g(t) = \cos t$ and $h(x) = x^2$. Both are continuous on $\mathbf{R}$ — one a trigonometric function, the other a polynomial — and the composition of continuous functions is continuous.

Question 32

Show that the function defined by $f(x) = |\cos x|$ is a continuous function.

Solution. Write $f = g \circ h$ with $g(t) = |t|$ and $h(x) = \cos x$. The modulus function is continuous everywhere (including at $0$, where it has a corner but no jump), and so is cosine. Hence the composition is continuous.

Question 33

Examine that $\sin|x|$ is a continuous function.

Solution. Here the composition is the other way round: $f = g \circ h$ with $g(t) = \sin t$ and $h(x) = |x|$. Both are continuous on $\mathbf{R}$, so $\sin|x|$ is continuous everywhere.

Question 34

Find all the points of discontinuity of $f$ defined by $f(x) = |x| – |x+1|$.

Solution. Each of $|x|$ and $|x+1|$ is continuous everywhere, and the difference of two continuous functions is continuous.

The corners at $x = 0$ and $x = -1$ are points where $f$ fails to be differentiable, not points where it fails to be continuous — a distinction the next exercise develops.

Common mistakes

• Calling a function discontinuous where it is undefined. In question 3(b), $x = 5$ is outside the domain; continuity is only ever asserted at points of the domain.
• Reading the wrong branch for $f(c)$. In question 15 the value at $x = 1$ comes from the middle branch ($\le 1$), giving $0$, not from $4x$.
• Confusing a corner with a break. Questions 3(d), 32 and 34 all have corners and are perfectly continuous.
• Assuming a parameter can always fix a discontinuity. In question 18 the left-hand limit is $0$ for every $\lambda$, so no value works.
• Evaluating a $\tfrac00$ form by substitution. Question 26 needs the substitution $x = \tfrac{\pi}{2} + h$ and the standard limit $\tfrac{\sin h}{h} \to 1$.
• Forgetting the sandwich theorem when a bounded factor oscillates. Question 24’s $\sin\tfrac1x$ has no limit at $0$, but $x^2\sin\tfrac1x$ does.

Practise next

• Exercise 5.2 — differentiability, and the chain rule for composite functions.
• Exercise 5.3 — implicit differentiation, once the machinery is in place.