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

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

Step 1: Check $f(a)$. From the definition, $f(a) = 0$.

Step 2: Evaluate $\lim_{x \to a} f(x)$. For $x \ne a$, we have $f(x) = (x - a) \sin \frac{1}{x - a}$. As $x \to a$, the factor $(x - a) \to 0$, while $\sin \frac{1}{x - a}$ oscillates between $-1$ and $1$ (bounded). By the Squeeze Theorem:

$-|x - a| \le (x - a) \sin \frac{1}{x - a} \le |x - a|$ …

Why This Problem Matters

Continuity at a point is a three-part test:

1. The function must be defined at the point.
2. The limit must exist as $x$ approaches that point.
3. The limit must equal the function value.

Here, the function is defined piecewise — a "normal" expression for $x \ne a$ and a separate value at $x = a$. The only potential trouble is whether the limit matches the assigned value. The trick is that the expression $(x - a) \sin \frac{1}{x - a}$ involves a sine of something that blows up, but it's multiplied by something that shrinks to zero. That's a classic squeeze-play situation.

Step-by-Step Solution

1. Check the function value at $x = a$

From the definition, $f(a) = 0$. So condition (1) is satisfied.

2. Examine the limit as $x \to a$

We need $\displaystyle \lim_{x \to a} f(x)$. For $x \ne a$, $f(x) = (x - a) \sin \frac{1}{x - a}$.

As $x \to a$, the factor $(x - a) \to 0$. The sine factor, however, oscillates wildly between $-1$ and $1$ because its argument $\frac{1}{x - a}$ goes to $\pm \infty$. So we cannot directly substitute.

3. Use the Squeeze Theorem

We know that for any real $t$, $-1 \le \sin t \le 1$. Therefore, for $x \ne a$:

Both the lower bound $-|x - a|$ and the upper bound $|x - a|$ approach $0$ as $x \to a$. …

Method: Constructing a Squeeze-Theorem Bound Around a General Point $x=a$

This is the general procedure for handling a limit of the form $(x-a)\cdot h\!\left(\frac{1}{x-a}\right)$, where $h$ is a bounded oscillating function (sine or cosine) — the same core idea as bounding-and-squeezing, but written so it applies at any point $a$, not just $x=0$.

Steps

Step 1: Isolate the vanishing factor and the oscillating factor.

Here the vanishing factor is $(x-a)$, which tends to $0$ as $x \to a$ by definition; the oscillating factor is $\sin\!\left(\frac{1}{x-a}\right)$, whose argument grows without bound as $x \to a$, so the factor itself has no limit.

Step 2: Use the universal bound on sine, $-1 \le \sin(\cdot) \le 1$, and translate it into a bound on the whole product using absolute values rather than a direct inequality (this avoids sign issues when $(x-a)$ can be negative).

Since $\left|\sin\left(\frac{1}{x-a}\right)\right| \le 1$ for all valid inputs, …

Common Mistakes

Mistake 1: Believing the limit fails to exist because $\sin\frac{1}{x-a}$ has no limit as $x\to a$.

Why it's wrong: exactly as with a bare oscillating factor, the vanishing factor $(x-a)$ tames the unbounded oscillation of $\sin\frac{1}{x-a}$; the product's limit exists and equals $0$ even though the sine factor by itself never settles. Correct approach: isolate the vanishing linear factor and bound the whole product using $|\sin(\cdot)|\le 1$, rather than trying to evaluate the oscillating part directly.

Mistake 2: Forgetting that the bound must use $|x-a|$, not $(x-a)$, since $(x-a)$ can be negative. …