Show that the function $$f(x)=\begin{cases}\dfrac{x^3-1}{x-1}, & x\ne1\\ 3, & x=1\end{cases}$$ is continuous on $(-\infty,\infty)$.

Show that the function f(x)= (x³-1)/(x-1), x≠1\3, x=1 is continuous on (-∞,∞).

Exercise 9.5 · Q5

f(x)=\begin{cases}\dfrac{x^3-1}{x-1}, & x\ne1\\ 3, & x=1\end{cases}

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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? …

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Split the domain into "away from $x=1$ " (automatic, by the algebra of continuous functions) and "at $x=1$ " (needs the factor-and-cancel trick), then combine both to cover all of $\mathbb R$ .

Step 1. Continuity for $x\ne1$ . For $x\ne1$ , $f(x)=\dfrac{x^3-1}{x-1}$ is a quotient of two polynomials (each continuous on $\mathbb R$ ) whose denominator $x-1$ is nonzero. By the algebra of continuous functions, a quotient is continuous wherever the denominator is nonzero, so $f$ is continuous at every $x\ne1$ .

Step 2. Continuity at $x=1$ — check $f(1)$ is defined. We are given $f(1)=3$ , so the first condition holds.

Step 3. Compute $\lim_{x\to1}f(x)$ . Factor the numerator using the difference-of-cubes identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x,\ b=1$ :

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

So for $x\ne1$ , $f(x)=\dfrac{(x-1)(x^2+x+1)}{x-1}=x^2+x+1$ . Hence

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

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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).

…

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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:

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

…

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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:

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

…

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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.

…

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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: $c=1-c\Rightarrow c=\dfrac12$ .

…

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

…

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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. …
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Q.Show that the function [!FORMULA] f(x)={x3−1x−1,x≠13,x=1f(x)=\begin{cases}\dfrac{x^3-1}{x-1}, & x\ne1\\ 3, & x=1\end{cases}f(x)=⎩⎨⎧​x−1x3−1​,3,​x=1x=1​ is continuous on (−∞,∞)(-\infty,\infty)(−∞,∞).

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Q.Show that the function
[!FORMULA] $f(x)=\begin{cases}\dfrac{x^3-1}{x-1}, & x\ne1\\ 3, & x=1\end{cases}$
is continuous on $(-\infty,\infty)$ .

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