How to calculate the range of $x\sin\frac{1}{x}$? I want to find the range of $f(x)=x\sin\frac{1}{x}$. It is clearly that its upper boundary is $$\lim_{x\to\infty}x\sin\frac{1}{x}=1$$ but what is its lower boundary? 
I used software to obtain the result $y\in[0.217234, 1]$ and the figure is

How to calculate the value '0.217234'? Thank you!

We could try to find the $x$ coordinate of that minimum using calculus.  Note that
$$
[x \sin (1/x)]' = \sin(1/x) - \cos(1/x)/x
$$
We would like to find the lowest $x$-value at which the derivative is zero.  That is, we want to find the lowest solution of 
$$
\sin(1/x) - \cos(1/x)/x = 0 \implies \\
\sin(1/x) = \cos(1/x)/x \implies\\
\tan(1/x) = 1/x
$$
Setting $y = 1/x$, what we're really doing is solving the equation $\tan(y) = y$.  Unfortunately, this equation can only be solved using numerical methods.  In this case, you would want the greatest negative solution $y$, and set $x = 1/y$.  That is, we want the solution $y = -4.49340945790906 \implies x = -0.222548158$
From there, you would calculuate $x \sin (1/x) = -0.217233628$.
