## Calculus 8th Edition

See The Squeeze Theorem, p.69: If $f(x) \leq g(x) \leq h(x)$, and $\displaystyle \lim_{x\rightarrow a}f(x)= \displaystyle \lim_{x\rightarrow a}h(x)=L,$ then $\displaystyle \lim_{x\rightarrow a}g(x)=L.$ ------ Let $g(x)=x^{2}\cos(1/x^{2})$ The range of cosine is $[-1,1],$ $-1 \leq \cos(1/x^{2}) \leq 1\qquad/\times x^{2}$ $-x^{2} \leq \cos(1/x^{2}) \leq x^{2}$ $\displaystyle \lim_{x\rightarrow 0}(-x^{2})=0$ and $\displaystyle \lim_{x\rightarrow 0}(x^{2})=0$ so, by the theorem, $\displaystyle \lim_{x\rightarrow 0}g(x)=0$