Q.Determine if $f$ defined by $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{if } x \ne 0 \\ 0, & \text{if } x = 0 \end{cases}$ is a continuous function?

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

We check continuity at $x = 0$ (the only possible trouble point).

Step 1: $f(0) = 0$ by definition.

Step 2: For $x \neq 0$, $|f(x)| = \left| x^2 \sin \frac{1}{x} \right| \le x^2$, since $|\sin(1/x)| \le 1$.

Step 3: As $x \to 0$, $x^2 \to 0$, so by the Squeeze Theorem, $\lim_{x \to 0} f(x) = 0$. …

The Core Idea: Continuity at a Point

Continuity at a point $x = a$ means three things must hold:

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

For this function, the only point where things could go wrong is $x = 0$, because that's where the definition changes. Everywhere else ($x \neq 0$), $f$ is a product of a polynomial ($x^2$) and a composition of continuous functions ($\sin(1/x)$ is continuous for $x \neq 0$). So continuity for $x \neq 0$ is immediate.

The real drama — and the beauty of this problem — is at $x = 0$. Here, $f(0) = 0$ is given. The question is: does $\lim_{x \to 0} x^2 \sin(1/x)$ exist, and if so, does it equal $0$?

Step-by-Step Solution

1. Recognize the "squeeze" situation.

The term $\sin(1/x)$ oscillates wildly as $x \to 0$ — it doesn't settle down to any single value. But it is bounded: for any $x \neq 0$,

This boundedness is the key. Multiply by $x^2$, which does approach $0$ as $x \to 0$. The product of something that goes to zero and something that stays bounded must also go to zero.

2. Set up the inequality.

For all $x \neq 0$,

3. Apply the Squeeze Theorem.

We know $\lim_{x \to 0} 0 = 0$ and $\lim_{x \to 0} x^2 = 0$. Since $|f(x)|$ is sandwiched between $0$ and $x^2$, the Squeeze Theorem gives:

4. From absolute value to the function itself. …

Method: Proving Continuity at a Point Using the Squeeze (Sandwich) Theorem

Use this when a function is defined piecewise with a factor like $\sin(1/x)$ or $\cos(1/x)$ that oscillates without settling to a single value as $x \to 0$, multiplied by a factor that shrinks to zero — direct substitution does not work, but bounding the oscillating part does.

Steps

Step 1: Recognise the obstruction

The function contains a term like $\sin(1/x)$ that has no limit as $x \to 0$ (it oscillates between $-1$ and $1$ infinitely often). You cannot evaluate $\lim_{x \to 0} f(x)$ by plugging in directly, because $\sin(1/x)$ itself isn't even defined at $x=0$.

Step 2: Bound the oscillating factor

Use the universal bound $-1 \le \sin(\text{anything}) \le 1$ to write

turning an unpredictable oscillation into a clean inequality.

Step 3: Apply the Squeeze Theorem

$0 \le |f(x)| \le x^2, \qquad \lim_{x\to 0} 0 = 0,\ \ \lim_{x \to 0} x^2 = 0 \implies \lim_{x \to 0}|f(x)| = 0$ …

Common Mistakes

Mistake 1: Concluding the limit doesn't exist because $\sin(1/x)$ itself has no limit

Why it's wrong: this ignores the effect of the $x^2$ factor multiplying it — a bounded oscillating quantity multiplied by something shrinking to zero is forced to zero, even though the oscillating factor alone never settles down. Correct approach: always check whether an oscillating term is being multiplied by a vanishing factor before declaring "no limit" — that's exactly when the Squeeze Theorem applies.

Mistake 2: Trying to compute $\lim_{x\to 0} \sin(1/x)$ directly instead of bounding it …