Answer
$$\lim_{x\to0}g(x)=0$$
Work Step by Step
$$g(x)=x\sin\Big(\frac{1}{x}\Big)$$
(a) The graph is shown below.
We see from the graph that as $x\to0$, the value of $g(x)$, though still oscillates, gets closer and closer to a single value $y$ of $0$.
This means we can estimate here that $\lim_{x\to0}g(x)=0$
(b) To prove our estimation, we cannot use normal methods or manipulate limit laws. We need to take advantage of the Sandwich Theorem.
We know that $-1\le\sin\Big(\frac{1}{x}\Big)\le1$
- For $x\ge0$, we have $-x\le x\sin\Big(\frac{1}{x}\Big)\le x$
- For $x\lt0$, we have $-x\ge x\sin\Big(\frac{1}{x}\Big)\ge x$
In any case, though, $g(x)=x\sin\Big(\frac{1}{x}\Big)$ is squeezed between $-x$ and $x$.
We have $\lim_{x\to0}x=\lim_{x\to0}(-x)=0$
Therefore, according to Sandwich Theorem, $\lim_{x\to0}g(x)=0$
