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!
 A: 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.
A: 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 $x \sin (1/x) = -0.217233628$.
A: 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.
