## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to0}x^2\cos(1/x^2)=0$$
$$A=\lim\limits_{x\to0}x^2\cos(1/x^2)$$ 1) We know that $$-1\le\cos(1/x^2)\le1$$ then $$-x^2\le x^2\cos(1/x^2)\le x^2$$ (since $x^2\ge0$, the inequality direction does not change) 2) Consider $\lim\limits_{x\to0}(-x^2)=-0^2=0$ and $\lim\limits_{x\to0}(x^2)=0^2=0$ So $\lim\limits_{x\to0}(-x^2)=\lim\limits_{x\to0}(x^2)=0$ 3) From 1) and 2), according to the Squeeze Theorem, we can conclude that $$\lim\limits_{x\to0}x^2\cos(1/x^2)=0$$