GATE 2023 IN Q.23: solution and answer

Official answer: ASource: Official answer key, GATE 2023 IN (IIT Kanpur)

1. As $x\to 0$, $\sin \frac{1}{x}$ keeps swinging between $-1$ and $1$, so it has no limit of its own. It is still bounded: $\left|\sin \frac{1}{x}\right|\le 1$ for every $x\ne 0$.

2. Multiply by $|x|$ to get a bound on $f$: $$0\le \left|x\sin \frac{1}{x}\right|=|x|\left|\sin \frac{1}{x}\right|\le |x|(x\ne 0)$$

3. Squeeze (sandwich) theorem: both bounds tend to 0, because ${\lim }_{x\to 0}0=0$ and ${\lim }_{x\to 0}|x|=0$. So ${\lim }_{x\to 0}\left|x\sin \frac{1}{x}\right|=0$.

4. Since $-|f(x)|\le f(x)\le |f(x)|$ and both outer terms tend to 0, the squeeze theorem gives $f(x)\to 0$ as well: $$\underset{x\to 0}{\lim }x\sin \frac{1}{x}=0$$

5. Put another way, a function that tends to 0 (here $x$) multiplied by a bounded function (here $\sin \frac{1}{x}$) tends to 0. The answer is (A).