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?

$$x \sin (1/x) =1 \implies \sin (\frac{1}{x})=\frac{1}{x} \implies \sin y =y.$$
But $$y > \sin y,~ if ~ y>0 ;~ y < \sin y ~ ~ if ~ ~ y<0.$$~
The only solution the last equation may have is $y=0$, but it is not allowed here as it would mean $x=\infty$. So the given equation cannot have a real root.
