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. 

Although an answer without derivatives and integrals is definitely more elegant, you can just use the fact that for $x>0$, $$\sin(x)=\int_0^x \cos y\,dy\leq \int_0^x 1\,dy= x$$ and similarly for $x<0$.
