Squeeze theorem applied to sin(x^2)/x

We are told that $-|x| \leq \dfrac{\sin(x^2)}{x} \leq |x|$ holds for all $x$ near $0$ (with $x \neq 0$). This sets up the conditions for the Squeeze Theorem, since our function of interest is bounded between two simpler functions.

Compute the limits of the bounding functions: $\lim_{x\to 0}(-|x|) = 0$ and $\lim_{x\to 0}|x| = 0$. Since both bounding functions approach the same value $0$ as $x \to 0$, the Squeeze Theorem guarantees that $\lim_{x\to 0} \dfrac{\sin(x^2)}{x} = 0$ as well.