Show that $|\frac {\sin x}{x}| < 1$ for all $ x \in \Bbb R $\ $\{0\}$ Problem : Show that $\left|\frac {\sin x}{x}\right| < 1$ for all $ x \in \Bbb R $\ $\{0\}$. 
The result seems obvious for $|x| \geq 1$, but I am not sure how to show it on the interval $(-1,0) \cup (0, 1)$. it seems this amounts to showing that $\sum_{n=0}^\infty \frac {|x^{2n}|}{(2n+1)!} < 1$, but I am not sure how to proceed, hints appreciated.
 A: 
For the case $x\in[-1,1]\setminus\{0\}$, consider a sector of a unit circle, with $x$ being the angle at the centre. The area is $$\frac12 |x|$$
Remove the circular segment from the sector, and what remains is a triangle. The area is
$$\frac12 |\sin x|$$
Since the triangle is just a part of the sector,
$$|\sin x| < |x|$$
A: For $0 \le x <1$, let
$f (x)=\sin (x)-x .$
$0 <x <1 \implies $
$f'(x)=\cos (x)-1 <0$
$\implies f $ strictly decreasing at $(0,1) $
$\implies \sin (x)-x < f (0)=0$
$\implies 0 \le \frac {\sin (x)}{x} <1$
observe that $f $ is odd.
A: Hint:
use the fact that $y=x$ is the line tangent to $y=\sin x$ at $x=0$ and $\sin x$ is concave down for $0< x< \pi$
A: For some $\xi$ between $0$ and $x$, the Mean Value Theorem says
$$
\frac{\sin(x)}{x}=\frac{\sin(x)-\sin(0)}{x-0}=\cos(\xi)\tag{1}
$$
Since $|\cos(\xi)|\lt1$ for all $\xi\ne0$ and $|\xi|\lt1$, $(1)$ implies that for $|x|\le1$
$$
\left|\frac{\sin(x)}{x}\right|\lt1\tag{2}
$$
Since $|\sin(x)|\le1$ for all $x$, we have $(2)$ for $|x|\gt1$, too. Thus, $(2)$ is true for all $x\ne0$.
A: Since it's even, you only need to consider $x>0$.
Because
$$\frac{d}{dx}\sin x = \cos x$$
so
$$1 > \frac{d}{dx}\sin x > -1$$
for $1>x>0$
Therefore
$$x > \sin x > -x$$
for $1>x>0$
