Is $-|x|\le\sin x\le|x|$ for all $x$ true? I have seen in Thomas' Calculus that says to prove $\lim_{x\rightarrow0}\sin x=0$, use the Sandwich Theorem and the inequality $-|x|\le\sin x\le|x|$ for all $x$.
My question is how could the inequality be true? If we derive from $-1\le\sin x\le1$, we could only get $-|x|\le|x|\sin x\le|x|$?
In the textbook it is written that it follows from the definition of $\sin x$ that $-|x|\le\sin x\le|x|$ for all $x$. Is there any easy way to show this?
Thanks for any help. 

The unit circle is parametrized by $\theta\mapsto(\cos\theta,\sin\theta)$:
$\hskip 1.5in$ 
Consider an angle $0\le\theta\le\frac{\pi}{2}$, as depicted above. The red line's length is $\theta$, and $\sin\theta$ is the purple length. Since they both travel the same vertical distance but the red one also travels horizontal distance, the red line is longer, hence $\sin\theta\le\theta$. (This can be proved formally with calculus if one so desires.) After $\theta>1$, the angle will forevermore be too much for sine to ever take for any angle. Since $\sin(-\theta)=-\sin\theta$, the claim follows for negative angles too.
