Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x \ne 0 \\ 5, & x = 0 \end{cases}$ at $x = 0$.

> [!TLDR] > The function is **discontinuous** at $x=0$ because the limit $\lim_{x\to 0} \frac{1-\cos 2x}{x^2} = 2$ does not equal the given function value $f(0)=5$. --- ## The Core Idea: Continuity at a Point A function is continuous at a point $x = a$ if three things match perfectly: 1. The function is defined at $a$ (we have $f(a)$). 2. The limit of $f(x)$ as $x$ approaches $a$ exists. 3. That limit equals $f(a)$. If any one of these fails — especially the third — the function is discontinuous there. Here, the function is defined piecewise: for $x \neq 0$, it's a trigonometric expression; at $x = 0$, it's simply $5$. The question is whether the behaviour of the function near $0$ (the limit) matches the value assigned at $0$. --- ## Step-by-Step Solution **1. Check the function value at $x = 0$** From the definition, $f(0) = 5$. That's straightforward. **2. Find $\lim_{x \to 0} f(x)$ for $x \neq 0$** We need to evaluate: $$ \lim_{x \to 0} \frac{1 - \cos 2x}{x^2} $$ Direct substitution gives $\frac{0}{0}$, an indeterminate form. So we must simplify. **3. Use a trigonometric identity** Recall the double-angle identity: $\cos 2x = 1 - 2\sin^2 x$. Then: $$ 1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x $$ So the expression becomes: $$ \frac{1 - \cos 2x}{x^2} = \frac{2\sin^2 x}{x^2} = 2 \cdot \left( \frac{\sin x}{x} \right)^2 $$ > [!TIP] > This identity is a classic shortcut — it turns a messy $\cos 2x$ limit into the standard $\frac{\sin x}{x}$ limit, which every exam expects you to know. **4. Apply the standard limit** We know: $$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $$ Therefore: $$ \lim_{x \to 0} 2 \left( \frac{\sin x}{x} \right)^2 = 2 \cdot (1)^2 = 2 $$ > [!WARNING] > A common mistake is to forget the square. Students sometimes write $\lim_{x\to 0} \frac{\sin x}{x} = 1$ but then drop the square when substituting. Always keep the exponent intact. **5. Compare the limit with the function value** We have: $$ \lim_{x \to 0} f(x) = 2 \quad \text{

Find whether the function is continuous or discont – Class 12 Mathematics

NCERT Exemplar · Q3

f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x \ne 0 \\ 5, & x = 0 \end{cases}

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Concept understanding — Continuity At A Point

Continuity at a Point

Imagine drawing the graph of a function and putting your pen down at $x = a$ . If the function is continuous there, you can draw straight through that point without lifting your pen — no jump, no hole, no break. That is the intuition; here is the precision.

The Three-Condition Test

For $f(x)$ to be continuous at $x = a$ , all three must hold. If even one fails, $f$ is discontinuous there.

Important

Continuity at $x = a$ requires:

$f(a)$ is defined,
$\displaystyle \lim_{x \to a} f(x)$ exists (left- and right-hand limits are equal),
$\displaystyle \lim_{x \to a} f(x) = f(a)$ .

Condition 1 says $a$ is in the domain — the pen must have somewhere to land. Condition 2 says the curve approaches a single value from both sides — no jump. Condition 3 says that common approach value actually matches the function's value at $a$ — no misplaced point.

Why All Three Are Needed

$f(x) = \dfrac{x^2 - 1}{x - 1}$ has $\lim_{x \to 1} f(x) = 2$ , yet $f(1)$ is undefined (zero denominator). Condition 1 fails, leaving a hole at $(1, 2)$ .

A piecewise function shows the opposite can be fine:

f(x) = \begin{cases} x + 1 & x < 2 \\ 3 & x = 2 \\ x + 1 & x > 2 \end{cases}

Here $f(2) = 3$ , both one-sided limits equal $3$ , and they match $f(2)$ — so all three hold and $f$ is continuous at $x = 2$ .

Common Pitfalls

Watch out

"Limit exists" does not mean "continuous." The hole example has a limit but no continuity — the limit must equal the function value.

Watch out

"Defined everywhere" does not mean "continuous." A piecewise function can have a value at every point and still jump. Always check the one-sided limits.

A Quick Check

Is $f(x) = |x|$ continuous at $x = 0$ ? $f(0) = 0$ , $\lim_{x \to 0}|x| = 0$ , and the two agree — yes, even though $|x|$ has a sharp corner. Continuity demands no break, not smoothness.

Takeaway: continuity at a point means the function value and the two one-sided limits all agree. Agreement $\Rightarrow$ the graph passes through unbroken; disagreement $\Rightarrow$ a discontinuity.

Continuity at a Point is the opening idea of the CBSE Class 12 Continuity and Differentiability chapter, and the three-condition test described here matches exactly what NCERT exercises and "continuity and differentiability class 12 important questions" expect students to apply. This concept is also foundational for JEE Main and NEET, where checking continuity is often the first step before testing differentiability of a function.

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Q.Find whether the function is continuous or discontinuous at the indicated point: f(x)={1−cos⁡2xx2,x≠05,x=0f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x \ne 0 \\ 5, & x = 0 \end{cases}f(x)=⎩⎨⎧​x21−cos2x​,5,​x=0x=0​ at x=0x = 0x=0.

Concept understanding — Continuity At A Point

Continuity at a Point

The Three-Condition Test

Why All Three Are Needed

Common Pitfalls

A Quick Check

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Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x \ne 0 \\ 5, & x = 0 \end{cases}$ at $x = 0$ .

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