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Why does $\underset{x\to {0}^{+}}{lim}\mathrm{s}\mathrm{g}\mathrm{n}\left(x{\mathrm{sin}}^{2}\left(\frac{1}{x}\right)\right)$ not exist?
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Shawn Castaneda
The limit doe not exist, because $f\left({x}_{n}\right)=0$ if $\phantom{\rule{thickmathspace}{0ex}}{x}_{n}=\frac{1}{n\pi }$ and $f\left(x\right)=1$ if $\phantom{\rule{thickmathspace}{0ex}}{x}_{n}=\frac{1}{\frac{\pi }{2}+n\pi }\phantom{\rule{1em}{0ex}}\left(n\in \mathbf{N}\right)$, and both sequences converge to 0