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?
 A: Notice that:
$$x \cdot \sin \left( \frac{1}{x} \right) = \frac{ \sin \left( \frac{1}{x} \right)}{\left( \frac{1}{x} \right)}.$$
Moreover, observe that:
$$\lim_{x \to +\infty} \frac{ \sin \left( \frac{1}{x} \right)}{\left( \frac{1}{x} \right)} = 1.$$
This means that your function "is equal to $1$" for $x \to +\infty$. Numerically, $x \to +\infty$ stands for a huge number. The solution provided by Wolfram ($x \approx 5.16\cdot 10^{15}$) is indeed a huge number.
A: 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.
A: $$x \sin (1/x) =1 \implies \sin (\frac{1}{x})=\frac{1}{x} \implies \sin y =y.$$
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.
