Squeeze theorem with a trigonometric bound

From $-x^2 \le h(x)(1-\cos x) \le x^2$, divide through by $1-\cos x>0$ (which is positive for small $x\ne 0$) to get $-\frac{x^2}{1-\cos x} \le h(x) \le \frac{x^2}{1-\cos x}.$ Multiplying every part of this inequality by $x\sin x$ preserves the inequality direction since $x\sin x \ge 0$ for $x$ near $0$. This gives $-\frac{x^3\sin x}{1-\cos x} \le x\,h(x)\sin x \le \frac{x^3\sin x}{1-\cos x}.$

To evaluate $\lim_{x\to 0} \frac{x^3\sin x}{1-\cos x}$, we use the given standard limits. Since $\lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2}$, we have $1-\cos x \sim \frac{x^2}{2}$ for small $x$. Therefore, $\left|\frac{x^3\sin x}{1-\cos x}\right| \approx \frac{x^3 \cdot |\sin x|}{x^2/2} \approx 2x|\sin x|.$ Since $|\sin x| \le 1$, we have $\left|\frac{x^3\sin x}{1-\cos x}\right| \le 2|x|$. As $x \to 0$, both bounds approach $0$. By the Squeeze Theorem, $\lim_{x\to 0} x\,h(x)\sin x = 0.$