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!

As you have been told in previous answers, the minimum will occur when $y-\tan (y)=0$ using $y=\frac{1}{x}$ that is to say when the straight line $z=y$ will intersect the curve $z=\tan(y)$. Overlaping the two graphs shows that the solution is somewhere close to $4.5$.
So, let us use Newton method with a guess $y_0=4.5$ which corresponds to the mid point of the interval. Newton iterations will update the guess using $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$ using $$f(y)=y-\tan (y)$$ $$f'(y)=1-\sec ^2(y)$$ So starting Newton scheme, the suceesive iterates are $4.49361$,  $4.49341$ which is the solution for six significant figures.
