Squeeze theorem applied to x^2 sin(1/x)

Since $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ holds for all $x \neq 0$, multiply every part of this inequality by $x^2 \ge 0$. This preserves the inequality direction and gives $-x^2 \le x^2\sin\left(\frac{1}{x}\right) \le x^2.$ So we have bounded $f(x) = x^2\sin(1/x)$ between $g(x) = -x^2$ and $h(x) = x^2$.

Now compute the limits of the bounding functions as $x \to 0$: $\lim_{x\to 0} (-x^2) = 0$ and $\lim_{x\to 0} x^2 = 0$. Since both bounds approach the same value $0$, the Squeeze Theorem guarantees that $f(x)$ is trapped between two functions converging to $0$, so $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0.$