Trig limit in Spivak's Calculus $$\lim_{x\rightarrow 1} (x-1)^3 \sin\frac{1}{(1-x)^3} = 0$$
To prove that this is true, the chapter on limits has things like $\lim_{x\rightarrow a}(f\cdot g)(x) = \lim_{x\rightarrow a}f(x)\cdot \lim_{x\rightarrow a} g(x)$, when both limits exist. Now, in the case of the problem, $\lim_{x\rightarrow 1} (x-1)^3 = 0$, but $\lim_{x\rightarrow 1} \sin\frac{1}{(1-x)^3}$ doesn't exist. However, it is bounded, and because the first limit tends to zero, it is clear that the multiplication tends to zero. But since there is a discontinuity where the limit doesn't exist, I can't find a way to express this formally in the terms of the theorems that I should know at this point. 
So there's my doubt, I guess. How should I proceed?

$$
\underbrace{-|x-1|^3}_A \quad \le \quad \underbrace{|x-1|^3\cdot\sin(\cdots)}_B \quad \le \quad \underbrace{|x-1|^3}_C.
$$
You don't need to know the nature of the function of $x$ inside the sine function in order to deduce the inequalities $A\le B\le C$.  If $A$ and $C$ both approach $0$, then so does $B$.  Most calculus textbooks have a theorem that states this explicitly.
