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!

It might be easier to replace $x$ by ${1 \over x}$... then your goal is to find the minimum of ${\sin x \over x}$. Taking derivatives, this occurs at an $x$ for which ${\cos x \over x} - {\sin x \over x^2} = 0$, or equivalently where $\tan x = x$. According to Wolfram Alpha, the first such minimum occurs at $x = 4.49340945790906\ldots$, corresponding to a value of ${\sin x \over x} = -0.217233628211222\ldots$. 
Since this is a transcendental equation you probably have to use numerical methods to find this value.. but even simple things like Newton-Raphson should work here.
