In Example 1.1.8, Principle 1.1.1 was used to establish $\underset{x\to 0}{lim}x \sin \left(1\/x\right)\=0$. Because $\left|\sin \left(1\/x\right)\right|\le 1$, and $x\to 0$, the limit is zero by Principle 1.1.1. The limit given in the present example would also yield to Principle 1.1.1 because $x^2\to 0$ also.

To establish the given limit via the Squeeze theorem, start with the following bound on $\sin \left(1\/x\right)$.

$-1\le \sin \left(1\/x\right)\le 1$ 

and multiply through by the positive quantity $x^2$. This gives

$-x^2\le x^2\sin \left(1\/x\right)\le x^2$ 

Since both $x^2$ and $-x^2$ go to zero, by the Squeeze theorem so also must $x^2\sin \left(1\/x\right)$ go to zero.