Squeeze theorem applied to |x|^(3/2) sin(1/x) as x → 0

Since $-1 \le \sin(1/x) \le 1$ for all $x \neq 0$, and $|x|^{3/2} \ge 0$, multiplying the inequality by $|x|^{3/2}$ preserves the direction of the inequalities: $-|x|^{3/2} \le |x|^{3/2}\sin\left(\frac{1}{x}\right) \le |x|^{3/2}.$ This gives us the squeeze bounds $g(x) = -|x|^{3/2}$ and $h(x) = |x|^{3/2}$.

Now evaluate the limits of the bounding functions: $\lim_{x\to 0} |x|^{3/2} = 0$ and $\lim_{x\to 0} -|x|^{3/2} = 0$, since raising $|x|$ to a positive power sends it to $0$ as $x \to 0$. Since both bounds tend to $0$, the Squeeze Theorem guarantees that $f(x)$ is trapped between two quantities both approaching $0$, so $\lim_{x \to 0} |x|^{3/2}\sin\left(\frac{1}{x}\right) = 0.$