Q.The function $f(x) = \dfrac{4 - x^2}{4x - x^3}$ is
(A) discontinuous at only one point
(B) discontinuous at exactly two points
(C) discontinuous at exactly three points
(D) none of these

The key idea is continuity at a point: a rational function is continuous wherever its denominator is non-zero. Discontinuities occur at points where the denominator vanishes, provided the numerator does not cancel the factor.

Step 1: Factor both numerator and denominator.

Step 2: Cancel the common factor $(x-2)(x+2)$, but note that cancellation is valid only when $x \neq \pm 2$. The simplified form is:

$f(x) = \frac{1}{x}, \quad x \neq 0, \pm 2$ …

The core idea: continuity at a point

A function $f$ is continuous at $x = a$ if three things hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists (finite).
3. $\lim_{x \to a} f(x) = f(a)$.

If any of these fails, $f$ is discontinuous at $a$. The tricky part here is that $f$ is given as a rational expression, and we must first find where it is defined — because if $f(a)$ is not defined, we already have a discontinuity. Then we check whether the discontinuity is removable (limit exists but function undefined) or non‑removable (limit does not exist or is infinite).

Step‑by‑step reasoning

1. Find the domain of the given expression

The function is $f(x) = \dfrac{4 - x^2}{4x - x^3}$. The denominator must not be zero:

So the denominator vanishes at $x = 0$, $x = 2$, and $x = -2$. At these three points, $f(x)$ is not defined. Therefore $f$ is automatically discontinuous at each of these points (condition 1 fails).

2. Simplify the expression where possible

Factor numerator and denominator:

Notice that $4 - x^2$ appears in both numerator and denominator. For $x$ such that $4 - x^2 \neq 0$ (i.e., $x \neq \pm 2$), we can cancel:

But at $x = \pm 2$, the factor $4 - x^2$ is zero, so cancellation is not allowed — the original expression becomes $0/0$, an indeterminate form. At $x = 0$, the denominator has a factor $x$ that does not cancel, so the expression blows up.

3. Analyse each point of discontinuity

• At $x = 2$: The original $f(2)$ is undefined. But consider the limit as $x \to 2$: for $x$ near 2 (but not equal to 2), we have $f(x) = 1/x$ (since $4 - x^2 \neq 0$ near 2). So

The limit exists and is finite. This is a removable discontinuity — if we redefined $f(2) = 1/2$, the function would become continuous at $x = 2$.

• At $x = -2$: Similarly, $f(-2)$ is undefined. For $x$ near $-2$ (but not equal to $-2$), $f(x) = 1/x$, so

Again the limit exists and is finite — another removable discontinuity.

• At $x = 0$: …

Method: Locating All Discontinuities of a Rational Function

Use this whenever a function is written as a ratio of two polynomials and you must find every point of discontinuity, including ones that 'disappear' after simplification.

Steps

Step 1: Factor the numerator and denominator completely, without cancelling anything yet.

Writing both in fully factored form reveals every root of the denominator, which is what actually determines the function's domain.

Step 2: Solve the original, uncancelled denominator $=0$.

Every root found here is a point where the given function is not defined — and by the continuity test, an undefined point is automatically a discontinuity, regardless of what happens after simplification.

Step 3: Cancel any common factors between numerator and denominator to get a simplified form, valid only where the cancelled factor is non-zero.

$\frac{(x-a)\,h(x)}{(x-a)\,k(x)} = \frac{h(x)}{k(x)}, \quad x \neq a$ …

Common Mistakes

Mistake 1: Simplifying $f(x)$ to $\frac{1}{x}$ first, then only checking the simplified expression for discontinuities.

Why it's wrong: this misses the points $x=\pm2$ entirely, because the factor $(4-x^2)$ that cancelled there was zero at exactly those points in the original function — the simplified form is only valid for $x \neq \pm 2$, so its own domain silently hides those discontinuities.

Correct approach: always find the zeros of the original, uncancelled denominator first, and treat every one of them as a discontinuity — only use the simplified form afterward, to classify each one as removable or not.

Mistake 2: Believing a removable discontinuity (a 'hole') doesn't count as a real discontinuity. …