Examine the continuity of the following: **(i)** $x+\sin x$ **(ii)** $x^2\cos x$ **(iii)** $e^x\tan x$ **(iv)** $e^{2x}+x^2$ **(v)** $x\ln x$ **(vi)** $\dfrac{\sin x}{x^2}$ **(vii)** $\dfrac{x^2-16}{x+4}$ **(viii)** $|x+2|+|x-1|$ **(ix)** $\dfrac{|x-2|}{|x+1|}$ **(x)** $\cot x+\tan x$

Examine the continuity of the following: **(i)** x+sin x **(ii)** x²cos x **(iii)** extan x **(iv)** e2x+x² **

Exercise 9.5 · Q2

x+\sin x

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Concept understanding — Continuity of a Function

Continuity of a Function

The Intuition: Drawing Without Lifting the Pen

Imagine you are drawing the graph of a function on a piece of paper. If you can trace the entire curve without lifting your pen from the paper, the function is continuous. Every time you have to lift the pen — because the graph jumps, breaks, or has a hole — the function is discontinuous at that point.

That is the visual idea. A continuous function has no sudden leaps, no gaps, no punctures. Its output changes smoothly as its input changes.

Consider a simple example: $f(x) = x^2$ . As $x$ moves from 1 to 2, the output moves from 1 to 4, passing through every value in between. No jump, no missing point. You can draw it in one stroke.

Now contrast that with a function like:

f(x) = \begin{cases} x^2 & \text{if } x \neq 1 \\ 5 & \text{if } x = 1 \end{cases}

At $x=1$ , the graph has a single isolated point at height 5, while the rest of the curve approaches height 1. To draw this, you would trace the parabola, then lift your pen to place a dot at $(1,5)$ . That lift is the discontinuity.

The Problem with Intuition Alone

"Drawing without lifting the pen" works for simple functions, but it fails for strange ones. Some functions are continuous yet impossible to draw (like the Weierstrass function, which is continuous but has no smooth tangent anywhere). More practically, the pen-lifting test is not a mathematical definition — it cannot tell you exactly what "no break" means at a single point.

We need a precise, point-by-point definition.

The Precise Definition: The Three-Part Test

A function $f(x)$ is said to be continuous at a point $x = a$ if and only if all three of the following conditions hold:

$f(a)$ is defined. The function must have a value at $x = a$ . No holes.
$\lim_{x \to a} f(x)$ exists. As $x$ gets arbitrarily close to $a$ from either side, the function's values must approach a single finite number.
$\lim_{x \to a} f(x) = f(a)$ . The limit must equal the actual function value. The point must sit exactly where the surrounding curve is heading.

If any one of these fails, the function is discontinuous at $x = a$ .

Important

Condition 3 is the heart of continuity. It says: "What the function should be (the limit) is exactly what it is (the value)." No surprises.

Why the Limit Matters

The limit captures the trend of the function near $a$ , ignoring what happens exactly at $a$ . Continuity demands that this trend matches the actual point. This is why the earlier piecewise function fails: the limit as $x \to 1$ is $1$ , but $f(1) = 5$ , so condition 3 is violated.

A Worked Example

Test whether $f(x) = \frac{x^2 - 1}{x - 1}$ is continuous at $x = 1$ .

Step 1: Is $f(1)$ defined?

No. The denominator becomes zero, so $f(1)$ is undefined. Condition 1 fails immediately. The function is discontinuous at $x = 1$ — it has a hole there.

Even though $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x+1) = 2$ , the function never actually takes the value 2 at $x=1$ . The hole is a discontinuity.

Continuity on an Interval

A function is continuous on an interval (like $(a,b)$ or $[a,b]$ ) if it is continuous at every point in that interval. For a closed interval $[a,b]$ , we only require one-sided continuity at the endpoints: from the right at $a$ , and from the left at $b$ .

The Big Picture

Continuity is the mathematical way of saying "no surprises." It guarantees that small changes in input produce small changes in output. This property is the foundation for everything that follows in calculus: the Intermediate Value Theorem, differentiability (every differentiable function is continuous), and the ability to evaluate limits by direct substitution for continuous functions.

Tip

For most functions you meet in school — polynomials, trigonometric functions, exponentials, logarithms — they are continuous everywhere on their natural domains. You only need to check points where the function is not defined (like division by zero) or where its rule changes (piecewise definitions).

Use the algebra-of-continuous-functions rules: sums and products of continuous functions are continuous everywhere both are defined; a quotient is continuous wherever the denominator is nonzero; a composition of continuous functions is continuous on its domain.

Step 1. Part (i) $x+\sin x$ . $x$ is a polynomial (continuous on $\mathbb R$ ) and $\sin x$ is a standard continuous function on $\mathbb R$ . Their sum is continuous on $\mathbb R$ — no exceptional points.

Step 2. Part (ii) $x^2\cos x$ . $x^2$ and $\cos x$ are each continuous on $\mathbb R$ ; a product of continuous functions is continuous. So $x^2\cos x$ is continuous on $\mathbb R$ .

Step 3. Part (iii) $e^x\tan x$ . $e^x$ is continuous on $\mathbb R$ . $\tan x=\sin x/\cos x$ is continuous wherever $\cos x\ne0$ , i.e. everywhere except $x=\tfrac\pi2+n\pi,\ n\in\mathbb Z$ (undefined there). The product is continuous on $\mathbb R\setminus\{\tfrac\pi2+n\pi\}$ ; discontinuous (undefined) at $x=\tfrac\pi2+n\pi$ .

Step 4. Part (iv) $e^{2x}+x^2$ . $e^{2x}$ (composition of $e^x$ with the linear/continuous $2x$ ) and $x^2$ are both continuous on $\mathbb R$ ; their sum is continuous on $\mathbb R$ .

Step 5. Part (v) $x\ln x$ . $\ln x$ is continuous only on its domain $(0,\infty)$ ; $x$ is continuous on $\mathbb R$ . The product $x\ln x$ is therefore continuous throughout its domain $(0,\infty)$ (it is simply not defined, so the question of continuity does not arise, for $x\le0$ ).

Step 6. Part (vi) $\dfrac{\sin x}{x^2}$ . $\sin x$ and $x^2$ are continuous on $\mathbb R$ ; the quotient is continuous wherever $x^2\ne0$ , i.e. everywhere except $x=0$ , where the function is undefined. So it is continuous on $\mathbb R\setminus\{0\}$ , discontinuous at $x=0$ .

Step 7. Part (vii) $\dfrac{x^2-16}{x+4}$ . Numerator and denominator are polynomials, continuous on $\mathbb R$ . The quotient is continuous wherever $x+4\ne0$ , i.e. for all $x\ne-4$ . At $x=-4$ the function is undefined, so it is discontinuous there. (In fact $\dfrac{x^2-16}{x+4}=\dfrac{(x-4)(x+4)}{x+4}=x-4$ for $x\ne-4$ , and $\lim_{x\to-4}(x-4)=-8$ exists, so this is a removable discontinuity — but the function as written is still discontinuous at $x=-4$ since it is undefined there.)

Step 8. Part (viii) $|x+2|+|x-1|$ . $|x+2|$ and $|x-1|$ are each compositions of the continuous absolute-value function with a linear (continuous) function, hence continuous on $\mathbb R$ . Their sum is continuous on $\mathbb R$ .

Step 9. Part (ix) $\dfrac{|x-2|}{|x+1|}$ . Numerator and denominator are each continuous on $\mathbb R$ (as in Step 8). The quotient is continuous wherever $|x+1|\ne0$ , i.e. everywhere except $x=-1$ . At $x=-1$ the numerator $|{-1}-2|=3\ne0$ while the denominator $\to0$ , so the function is undefined and, in fact, unbounded near $x=-1$ — an infinite (non-removable) discontinuity there.

Step 10. Part (x) $\cot x+\tan x$ . $\cot x=\cos x/\sin x$ is continuous except where $\sin x=0$ , i.e. $x=n\pi$ . $\tan x=\sin x/\cos x$ is continuous except where $\cos x=0$ , i.e. $x=\tfrac\pi2+n\pi$ . Together, $\{n\pi\}\cup\{\tfrac\pi2+n\pi\}=\{\tfrac{n\pi}2:n\in\mathbb Z\}$ . The sum $\cot x+\tan x$ is continuous everywhere except at $x=\tfrac{n\pi}2,\ n\in\mathbb Z$ .

✓Final answer

x+\sin x

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$f(x)=\dfrac{1}{x}$
(a) $(-\infty, 0]$
(b) $\mathbb{R}$
(c) $[0, \infty)$
(d) $\mathbb{R}-\{0\}$
›Reveal solutionSolution

$f(x)=1/x$ has domain $\mathbb{R}-\{0\}$ (undefined at $x=0$ ), and it is continuous at every point of this domain.

$f(x)=\dfrac{1}{x}$ is a rational function, defined for all $x\ne0$ .

Rational functions are continuous at every point of their domain (quotient of continuous polynomials, valid wherever the denominator is nonzero).

Since $x=0$ is not in the domain at all, continuity there is not even a meaningful question — $f$ is continuous throughout $\mathbb{R}-\{0\}$ .

✓Final answer
$\mathbb{R}-\{0\}$
CBSE 2026Set SEM31 markMCQ
$f(x) = \dfrac{x^2 - 1}{x^3 - 1}$
(a) $1$
(b) $\dfrac{1}{3}$
(c) $\dfrac{2}{3}$
(d) $2$
›Reveal solutionSolution

Cancel the common $(x-1)$ factor and take the limit; continuity forces $f(1)$ to equal that limit, $\dfrac23$ .

Removable discontinuity and continuity at a point is a CBSE/NCERT Class 12 continuity topic.

Factor numerator and denominator:

$f(x) = \frac{x^2-1}{x^3-1} = \frac{(x-1)(x+1)}{(x-1)(x^2+x+1)} = \frac{x+1}{x^2+x+1}, \quad x\neq 1.$

For $f$ to be continuous at $x=1$ we set $f(1)$ equal to the limit:

$f(1) = \lim_{x\to1}\frac{x+1}{x^2+x+1} = \frac{2}{1+1+1} = \frac{2}{3}.$

✓Final answer
$f(1) = \dfrac{2}{3}$
CBSE 2026Set SEM31 markMCQ
$f(x) = \dfrac{x^2 + 4x + 3}{x^3 + 3x^2 - x - 3}$
(a) $x = 1, -1, -3$
(b) $x = -1, -3$
(c) $x = 1, -3$
(d) $x = 1, -1$
›Reveal solutionSolution

$f$ is discontinuous where its denominator is zero; factor $x^3+3x^2-x-3$ to find those points.

Points of discontinuity of a rational function occur where the denominator vanishes — a CBSE/NCERT Class 12 continuity topic.

Factor the denominator by grouping:

$x^3 + 3x^2 - x - 3 = x^2(x+3) - (x+3) = (x+3)(x^2-1) = (x+3)(x-1)(x+1).$

The function $f(x)=\dfrac{x^2+4x+3}{(x+3)(x-1)(x+1)}$ is undefined (hence discontinuous) wherever the denominator is zero:

$x = 1,\ -1,\ -3.$

(Note: the numerator $x^2+4x+3=(x+1)(x+3)$ cancels common factors, making $x=-1$ and $x=-3$ removable discontinuities and $x=1$ an infinite discontinuity — but all three are genuine points of discontinuity of $f$ .)

✓Final answer
$x = 1, -1, -3$
CBSE 2022Set ANNUAL1 markMCQ
$x = \frac{3}{2}$
(a) differentiable
(b) continuous
(c) non-zero
(d) discontinuous
›Reveal solutionSolution

f(x) = |2x-3|/(2x-3) is the sign function of (2x-3); it is undefined at x = 3/2 and jumps from -1 to +1 there, so it is discontinuous at x = 3/2.

For $x > \frac{3}{2}$ , $2x-3>0$ , so $|2x-3|=2x-3$ and $f(x) = \frac{2x-3}{2x-3} = 1$ .

For $x < \frac{3}{2}$ , $2x-3<0$ , so $|2x-3|=-(2x-3)$ and $f(x) = \frac{-(2x-3)}{2x-3} = -1$ .

At $x=\frac{3}{2}$ itself, the denominator $2x-3=0$ , so $f(x)$ is not even defined there.

Since the left-hand limit is $-1$ , the right-hand limit is $+1$ , and the function value doesn't exist, $f$ is discontinuous at $x=\frac{3}{2}$ (and therefore also not differentiable there, since differentiability requires continuity).

✓Final answer
$The correct option is (d) discontinuous.$
CBSE 2020Set ANNUAL1 markMCQ
$f: \mathbf{R} \to \mathbf{R}$
(a) discontinuous at $x = \dfrac{1}{2}$
(b) continuous at $x = \dfrac{1}{2}$
(c) continuous everywhere
(d) discontinuous everywhere
›Reveal solutionSolution

This function is continuous at exactly the one point where its two branches meet, $x=\dfrac12$ .

$f(x)=x$ for irrational $x$ and $f(x)=1-x$ for rational $x$ . Near any point $x=c$ , both rationals and irrationals arbitrarily close to $c$ exist, so for $f$ to be continuous at $c$ we need the two branch formulas to agree there: $c=1-c\Rightarrow c=\dfrac12$ .

At $c=\dfrac12$ : whether we approach along rationals (giving values $\to1-\tfrac12=\tfrac12$ ) or irrationals (giving values $\to\tfrac12$ ), the limit is $\tfrac12$ , and $f(\tfrac12)=1-\tfrac12=\tfrac12$ (since $\tfrac12$ is rational) — limit equals function value, so $f$ IS continuous at $x=\tfrac12$ .

At any other point $c\ne\tfrac12$ , $c\ne1-c$ , so approaching along rationals gives one limiting value and along irrationals gives a different one — $f$ is discontinuous there.

✓Final answer
$x=\dfrac12$
CBSE 2019Set ANNUAL1 markMCQ
$f(x) = (1+2x)^{\frac{1}{x}}$
(a) e
(b) $e^2$
(c) 0
(d) 2
›Reveal solutionSolution

Use the standard limit $\displaystyle\lim_{x\to0}(1+ax)^{1/x}=e^a$ .

For continuity at $x=0$ : $f(0) = \displaystyle\lim_{x\to0}(1+2x)^{1/x}$

Using the standard limit $\displaystyle\lim_{x\to0}(1+ax)^{1/x} = e^{a}$ with $a=2$ :

$f(0) = e^2$

✓Final answer
$e^2$
CBSE 2018Set ANNUAL1 markMCQ
$f(x) = \tan x$
(a) $\left[\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right]$
(b) $(-\infty, \infty)$
(c) $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$
(d) $\left[0, \dfrac{\pi}{2}\right]$
›Reveal solutionSolution
$\tan x$ is continuous everywhere it is defined; it is undefined exactly at odd multiples of $\pi/2$ , so among the given intervals only the open interval $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ avoids those points entirely.

$\tan x = \frac{\sin x}{\cos x}$ is undefined wherever $\cos x = 0$ , i.e. at $x = \pm\pi/2, \pm3\pi/2,\ldots$

$\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ — closed interval, includes the undefined endpoints $\pm\pi/2$ . Not valid.

$(-\infty,\infty)$ — includes infinitely many undefined points. Not valid.

$\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ — open interval, both endpoints excluded, so $\tan x$ is defined and continuous throughout. Valid.

$\left[0,\frac{\pi}{2}\right]$ — closed, includes the undefined point $\pi/2$ . Not valid.

✓Final answer
$\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$

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