<s>[INST] Does $x \cdot \sin \left( \frac{1}{x} \right) = 1$ has solutions? A look to the plot show to you that the function $f(x) = x \cdot \sin \left( \frac{1}{x} \right) - 1 $ has no zero near the origin. Wolframalpha software says that $ x \cdot \sin \left( \frac{1}{x} \right) = 1$ if $x \approx 5.16\cdot 10^{15}$ but I suspect this huge number is a wrong solution due to numerical problems. Is it true that for some $x$, far away from the origin, the equation $$  x \cdot \sin \left( \frac{1}{x} \right) = 1 $$ is satisfied? Can, far away, $f(x)= x \cdot \sin \left( \frac{1}{x} \right)$ exceeds 1?
 [/INST] Consider substituting \frac{1}{x}: 
$$
x \cdot \sin \left( \frac{1}{x} \right) = 1\\
\iff\\
\sin \left( \frac{1}{x} \right) = \frac{1}{x}\\
\iff\\
\sin (z) = z\\
$$
This is only satisfied if $z=0$, implying $x=\infty$, which is what wolfram was trying to say.
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