Answer
$$1$$
Work Step by Step
Since
$$
\lim _{x \rightarrow -\infty} x \sin \frac{1 }{x}=-\infty\times0
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follws
$$
\lim _{x \rightarrow -\infty} x\sin \frac{1 }{x}=\lim _{x \rightarrow -\infty} \frac{\sin \frac{1 }{x}}{1/x}=\lim _{x \rightarrow -\infty} \frac{-\frac{1}{x^2}\cos \frac{1 }{x}}{-\frac{1}{x^2}}=\lim _{x \rightarrow -\infty} \cos \frac{1 }{x}=1.
$$
