How to find the absolute minimum of $f(x)=\frac{\sin{x}}{x}$ $$f(x)=\frac{\sin{x}}{x}$$
My calculusbook states that because $|f(x)|<\frac{1}{|x|}$ therefore the absolute minima of $f(x)$ must be at the two minima closest to the origin.
Please help me see why this is. I do not understand why the conclusion follows... I could also see it possible that another minimum would be the absolute minimum...
 A: I think your calculus book is a bit fast on this. But for $x\in (\pi,2\pi)$ (where $\sin$ is negative) and $k\geq 1$ you have:
$$ \frac{\sin(x+ 2\pi k)}{x+2\pi k} =\frac{\sin x}{x+2\pi k} > \frac{\sin x}{x}$$
which implies the claim.
A: One could just do the math, but you seem to ask for the intuition-part, so that will be my focus:
Think of $f(x)$ as the product of $\sin(x)$ and $1/x$.
The closer to $x=0$ you get, the less the sine is "dampened" by $1/x$. So the closer to $x=0$, the greater are the values that $f(x)$ can possibly attain. Thus the minima of the sine closest to $x=0$ will be the global minima of $f(x)$. 
Here's a plot: WA
A: All local minima besides those two are at a distance exceeding $3\pi$ from $0$.  So at those points we have $$\frac{\sin x} x \ge \frac{-1}{3\pi}.$$ The two critical points nearest the origin are at the points slightly less (in absolute value) than $3\pi/2$ at which $\dfrac{\sin x} x = \cos x.$ Maybe there's a slick way to proceed from here, but brute force suffices: do a bit of number-crunching to see that the value of $\dfrac{\sin x} x$ at that point is less than $-1/(3\pi)$.
Fun fact: the sum of the squares of the reciprocals of the positive values of $x$ that are critical points of $\dfrac{\sin x} x$ is exactly $1/10$.
A: Notice that $f$ is even. So $f(-x) = f(x)$. Then, notice that as $x \to \infty$, the values of $1/|x|$ become arbitrarily small. So too does $f$. However, if you look at $x < 0$, since $1/|x|$ decreases with decreasing $x$, the first minimum will be the smallest of all subsequent minima. Since $f$ is even, its largest minima are the first minima that are placed symmetrically about $x = 0$.
