vanhees71 said:7e: I guess everything is within real calculus. Then you can argue that the sine is bound and thus the limit must be zero:
0 \leq |x \sin(1/x)| \leq |x| \rightarrow 0.
7g: Expanding \sin(1/x) around x \rightarrow \infty gives
x \sin(1/x)=x [1/x+\mathcal{O}(1/x^3)] \rightarrow 1.