[tex]\bf \textit{we know the range for }cos\left( \frac{1}{x} \right)\textit{ is }[-1,1]\textit{ therefore}
\\\\\\
-1~\ \textless \ ~cos\left( \frac{1}{x} \right)~\ \textless \ ~1\impliedby \textit{multiplying all sides by }x^2
\\\\\\
-1x^2~\ \textless \ ~x^2cos\left( \frac{1}{x} \right)~\ \textless \ ~1x^2\implies -x^2~\ \textless \ ~x^2cos\left( \frac{1}{x} \right)~\ \textless \ ~x^2[/tex]
if the limit of -x² goes to "something", and the limit of x² goes to the same "something", if their limit coincide, and yet they're bounding the cosine expression, therefore, since the cosine expression is "sandwiched" between -x² and x², then the cosine expression "squeezes in" that little sliver between both -x² and x², and will inevitably go to the same limit.
[tex]\bf \begin{array}{ccccc}
\lim\limits_{x\to 0} -x^2&\ \textless \ &x^2cos\left( \frac{1}{x} \right)&\ \textless \ &\lim\limits_{x\to 0} x^2\\
\downarrow &&&&\downarrow \\
0&&&&0\\
&&\lim\limits_{x\to 0}x^2cos\left( \frac{1}{x} \right)\\
&&\downarrow \\
&&0
\end{array}[/tex]
