How do you use the Squeeze Theorem to find \[\mathop {\lim }\limits_{} ({x^2})cos20\left( {\pi \cdot x} \right)\] as $x$ approaches zero?

Hint: For solving this particular question we will consider a trigonometry concept that is range of the cosine function, \[ - 1 \leqslant cos\theta \leqslant 1\] where \[\theta \] is representing set of real numbers. we have to consider two limits right hand limit that is when \[x > 0\] , left hand limit that is when \[x < 0\] . After finding the corresponding limits we will conclude our result accordingly.

Complete step-by-step answer:
From trigonometry we have the range for the cosine function as
  \[ - 1 \leqslant cos\theta \leqslant 1\]
where \[\theta \] is representing a set of real numbers.
So, we can say that ,
 \[ - 1 \leqslant \mathop {\lim }\limits_{} ({x^2})cos20\left( {\pi \cdot x} \right) \leqslant 1\] for all $x$is not equal to zero.
Taking right hand limit,
Here \[x > 0\] , we will multiply without changing the directions of the inequalities, so we will get the following :
 \[ - {x^2} \leqslant \mathop {\lim }\limits_{} ({x^2})cos20\left( {\pi \cdot x} \right) \leqslant {x^2}\] for \[x > 0\] .
Taking , $\mathop {\lim }\limits_{x \to {0^ + }} \left( { - {x^2}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{x^2}} \right) = 0$
so, $\mathop {\lim }\limits_{x \to {0^ + }} ({x^2})cos20\left( {\pi \cdot x} \right) = 0$
Taking left hand limit,
Here \[x < 0\] , we will multiply, and we must change the directions of the inequalities, so we will get:
 \[ - {x^2} \leqslant \mathop {\lim }\limits_{} ({x^2})cos20\left( {\pi \cdot x} \right) \leqslant {x^2}\] for \[x < 0\] .
Taking , $\mathop {\lim }\limits_{x \to {0^ - }} \left( { - {x^2}} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {{x^2}} \right) = 0$,
so, $\mathop {\lim }\limits_{x \to {0^ + }} ({x^2})cos20\left( {\pi \cdot x} \right) = 0$
After calculating two-sided Limit , we conclude that:
$\mathop {\lim }\limits_{x \to 0} ({x^2})cos20\left( {\pi \cdot x} \right) = 0$ Because both the left-hand and right-hand limits are zero.

Note: While solving the left-hand limit or right-hand limit we can get the indeterminate form, there we will use L'Hôpital's Rule.
L'Hôpital's Rule says:
If you have got an indeterminate form for your answer to your limit, then you'll take the derivative of the numerator and of the denominator separately so as to search out the limit.
You can repeat this process if you still get an indeterminate form. An indeterminate form is when the limit seems to approach into a deeply weird answer.