#### Answer

please see step-by-step.

#### Work Step by Step

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 $\displaystyle \lim_{x\rightarrow a}g(x)=L.$
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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,
$\displaystyle \lim_{x\rightarrow 0}g(x)=0$