Question

Could some explain the sandwich theorem (calculus/ limits)

Answer 1

$\bf \textit{we know the range for }cos\left( \frac{1}{x} \right)\textit{ is }[-1,1]\textit{ therefore} \\\\\\ -1~\ \textless \ ~cos\left( \frac{1}{x} \right)~\ \textless \ ~1\impliedby \textit{multiplying all sides by }x^2 \\\\\\ -1x^2~\ \textless \ ~x^2cos\left( \frac{1}{x} \right)~\ \textless \ ~1x^2\implies -x^2~\ \textless \ ~x^2cos\left( \frac{1}{x} \right)~\ \textless \ ~x^2$

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.

$\bf \begin{array}{ccccc} \lim\limits_{x\to 0} -x^2&\ \textless \ &x^2cos\left( \frac{1}{x} \right)&\ \textless \ &\lim\limits_{x\to 0} x^2\\ \downarrow &&&&\downarrow \\ 0&&&&0\\ &&\lim\limits_{x\to 0}x^2cos\left( \frac{1}{x} \right)\\ &&\downarrow \\ &&0 \end{array}$