Infimum of $\frac{\sin x}{x}$ over $\mathbb{R}^+$ How do I find $\inf$ and $\sup$   of $\frac{\sin(x)}{x}$ for  $x\gt0$ ?
IMO it's quite easy to guess, but since I am on Math studies everything needs to be formal and I am not allowed to use things like derivatives just yet.
Any help or hints will be appreciated.
 A: For any $x>0$ we have 
$\sin x < x < \tan x$ a
as $\sin x>0$ we can divide the previous inequality by $\sin x$ and get
$$1 < \frac{x}{\sin x}< \frac{1}{\cos x}$$
reversing the fractions we have
$$\cos x < \frac{\sin x}{x} < 1$$
therefore as the minimum of $\cos x$ is $-1$ we have
$$-1<\frac{\sin x}{x}<1$$
So $1$ is clearly a sup, never attained on the function domain.
The inequality can be also taken from another point of view
we have $-1\le \sin x\le +1$ and therefore For any $x>0$ 
$$-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$$
Therefore the minimum of $\frac{\sin x}{x}$ is close to the $x$ where
$$\frac{\sin x}{x}=-\frac{1}{x}$$
that is when $\sin x = -1\to x=\frac{3\pi}{2}$
$$\frac{\sin{\frac{3\pi}{2}}}{{\frac{3\pi}{2}}}=-\frac{2}{3\pi}\approx -0.2122$$
Look the image at the bottom. The actual minimum is about $\color{red}{-0.217234}$ and is different because of course at the minimum the tangent is horizontal and no point on the hyperbola has an horizontal tangent. The actual minimum can be found setting zero the derivative of the function.
Hope this helps
$$...$$


