Chapter 1.6, Problem 49ES

In Example 3 we used the Squeezing Theorem to prove that $\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)=0$ Why couldn’t we have obtained the same result by writing $\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)=\underset{x\to 0}{\lim }x\cdot \underset{x\to 0}{\lim }\sin \left(\frac{1}{x}\right)=0\cdot \underset{x\to 0}{\lim }\sin \left(\frac{1}{x}\right)=0$ ?