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 $x = 0$.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: Compute the limit as $x \to 0$.

Use the identity $1 - \cos 2x = 2\sin^2 x$:

Step 2: Since $\lim_{x \to 0} \frac{\sin x}{x} = 1$, we get:

Step 3: Compare with $f(0) = 5$.

Since $\lim_{x \to 0} f(x) = 2 \neq 5 = f(0)$, the function is not continuous at $x = 0$.

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:

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:

So the expression becomes:

4. Apply the standard limit

We know:

Therefore:

5. Compare the limit with the function value

We have:

Since $2 \neq 5$, the third condition for continuity fails.

Method: Resolving a Trigonometric 0/0 Limit to Test Continuity

This method applies whenever the continuity check at a point reduces to a limit that hits the indeterminate form $\frac{0}{0}$ built from trigonometric expressions like $1-\cos\theta$ or $\sin\theta$.

Steps

Step 1: Check what's already given, and isolate what needs computing.

In a piecewise definition, the value at the point (here $f(0)$) is usually stated directly — note it down first. The real work is computing $\displaystyle\lim_{x\to a} f(x)$ for the "$x \ne a$" branch.

Step 2: Detect the indeterminate form and choose the right identity.

Direct substitution into a trig expression involving $1-\cos(\cdot)$ or $\sin(\cdot)$ over a power of $x$ typically gives $\frac{0}{0}$. The standard move is the double-angle identity

which converts the expression into a form built entirely from $\sin(\cdot)$, ready for the standard limit.

Step 3: Apply the standard limit $\displaystyle\lim_{t\to 0}\frac{\sin t}{t}=1$.

Rewrite the simplified expression as a product/quotient of $\dfrac{\sin t}{t}$-type pieces (possibly squared, if the identity introduced a square), and substitute the value $1$ for each such factor. Keep track of any exponent introduced by the identity — dropping a square here is the single most common slip.

Step 4: Compare the computed limit with the given function value.

If $\displaystyle\lim_{x\to a} f(x) = f(a)$, the function is continuous at $a$; if they differ, it is discontinuous there — no further checking is needed once a mismatch (or a non-existent limit) is found.

Common Mistakes

Mistake 1: Dropping the square when applying the standard limit.

Why it's wrong: after using $1-\cos 2x = 2\sin^2 x$, the expression becomes $2\left(\dfrac{\sin x}{x}\right)^2$ — the square is part of the algebra, not optional. Substituting $\lim \frac{\sin x}{x}=1$ but forgetting to square it gives an answer that's off by an incorrect factor. Correct approach: keep the square through every step and only substitute the numeric value $1$ at the very end, after squaring.

Mistake 2: Assuming continuity just because the limit exists.

Why it's wrong: here $\lim_{x\to 0} f(x) = 2$ genuinely exists, but the problem's whole point is that it does not equal $f(0)=5$. A student who stops at "the limit exists" without comparing it to the given value at the point will wrongly conclude continuity. Correct approach: always complete the third condition — compare the computed limit to the function's actual assigned value at that point.