Given information
We have to prove that the limit: \[\mathop {\lim }\limits_{x \to 0} {x^4}\cos \dfrac{2}{x} = 0\]We will use squeeze theorem to find the above limit:
According to the theorem, If $f\left( x \right) \le g\left( x \right) \le h\left( x \right)$ and for some a near $x$, $\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} h\left( x \right) = L$, then $\mathop {\lim }\limits_{x \to a} g\left( x \right) = L$

Step 1:
We know that cosine of any number lies between -1 and 1: \[ - 1 \le \cos \left( {\dfrac{2}{x}} \right) \le 1\]Since, $x^4$ is a positive number, multiplying the inequality both sides will have no effect \[\begin{array}{l} - 1 \le \cos \left( {\dfrac{2}{x}} \right) \le 1\\ - 1 \times {x^4} \le {x^4} \times \cos \left( {\dfrac{2}{x}} \right) \le 1 \times {x^4}\\ - {x^4} \le {x^4}\cos \left( {\dfrac{2}{x}} \right) \le {x^4}\end{array}\]

Step 2:
In the above inequality, let: \[f\left( x \right) = - {x^4};\,\,\,g\left( x \right) = {x^4}\cos \left( {\dfrac{2}{x}} \right);\,\,h\left( x \right)\,\, = {x^4}\]Now find the limit of $f(x)$ and $g(x)$: \[\begin{array}{l}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( { - {x^4}} \right)\\\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \left( { - {0^4}} \right)\\\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 0\end{array}\]\[\begin{array}{l}\mathop {\lim }\limits_{x \to 0} h\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( {{x^4}} \right)\\\mathop {\lim }\limits_{x \to 0} h\left( x \right) = \left( {{0^4}} \right)\\\mathop {\lim }\limits_{x \to 0} h\left( x \right) = 0\end{array}\]So using the squeeze theorem: \[\mathop {\lim }\limits_{x \to 0} g\left( x \right) = 0\]