haha1234
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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:
[tex]0 \leq |x \sin(1/x)| \leq |x| \rightarrow 0.[/tex]
7g: Expanding [itex]\sin(1/x)[/itex] around [itex]x \rightarrow \infty[/itex] gives
[tex]x \sin(1/x)=x [1/x+\mathcal{O}(1/x^3)] \rightarrow 1.[/tex]