Q.If $f(x) = x^2 \sin \dfrac{1}{x}$, where $x \ne 0$, then the value of the function $f$ at $x = 0$, so that the function is continuous at $x = 0$, is
(A) $0$
(B) $-1$
(C) $1$
(D) none of these

The key idea is continuity at a point: for $f$ to be continuous at $x=0$, we must have $\lim_{x\to 0} f(x) = f(0)$.

We are given $f(x) = x^2 \sin \frac{1}{x}$ for $x \neq 0$. To find the required $f(0)$, evaluate the limit:

1. For $x \neq 0$, $\sin \frac{1}{x}$ oscillates between $-1$ and $1$, so $-x^2 \le x^2 \sin \frac{1}{x} \le x^2$. …

The core idea here is continuity at a point. A function $f$ is continuous at $x = a$ if three things hold: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and they are equal. Here, the function is given as $f(x) = x^2 \sin \frac{1}{x}$ for $x \neq 0$, and we are free to choose $f(0)$ so that continuity holds. That means we must find what $\lim_{x \to 0} f(x)$ equals, and then set $f(0)$ to that value.

The tricky part is the $\sin(1/x)$ term. As $x$ approaches $0$, $1/x$ blows up to infinity, so $\sin(1/x)$ oscillates wildly between $-1$ and $1$. It does not have a limit. But notice: $x^2$ is shrinking to $0$. So we have a product: something that oscillates but stays bounded, multiplied by something that goes to zero. That is a classic situation for the Squeeze Theorem.

Let’s work through it step by step.

1. Set up the inequality. For any real number $t$, we know $-1 \le \sin t \le 1$. Here $t = 1/x$, so for all $x \neq 0$:

2. Multiply through by $x^2$. Since $x^2$ is always non-negative (and positive for $x \neq 0$), multiplying an inequality by a positive number preserves the direction:

3. Take limits as $x \to 0$. The left and right functions both have simple limits:

4. Apply the Squeeze Theorem. Since $f(x) = x^2 \sin(1/x)$ is trapped between two functions that both approach $0$, the squeeze theorem tells us: $\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0$ …

Method: Defining a Function at a Point via the Squeeze Theorem

Use this whenever you must choose the value $f(a)$ that makes a function continuous at $x=a$, and the formula for $x \neq a$ contains a bounded but non-convergent oscillating factor (like $\sin$ or $\cos$ of something blowing up).

Steps

Step 1: Identify the structure as (something $\to 0$) $\times$ (something bounded but with no limit).

Here the vanishing factor is $x^2 \to 0$, and the bounded oscillating factor is $\sin\frac{1}{x}$, which never settles as $x \to 0$ but always stays between $-1$ and $1$.

Step 2: Write the boundedness as a two-sided inequality.

Step 3: Multiply through by the vanishing factor, being careful with the direction of the inequality (multiplying by a positive quantity like $x^2$ keeps the direction unchanged). …

Common Mistakes

Mistake 1: Concluding that $\lim_{x\to 0} x^2\sin\frac{1}{x}$ does not exist because $\sin\frac{1}{x}$ itself has no limit as $x\to 0$.

Why it's wrong: this ignores that the oscillating factor is bounded (always between $-1$ and $1$) while the other factor $x^2$ shrinks to zero — a bounded quantity times something going to zero always goes to zero, regardless of how the bounded part behaves.

Correct approach: apply the Squeeze Theorem — bound $\sin\frac{1}{x}$ between $-1$ and $1$, multiply through by $x^2$, and take the limit of the two simple outer bounds, both of which clearly go to $0$. …