Answer
$$\frac{1}{{{e^2}}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} = {\left[ {\cos \left( {\frac{2}{{ + \infty }}} \right)} \right]^{{{\left( { + \infty } \right)}^2}}} = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }{\text{ }} \cr
& {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} = {e^{{x^2}\ln \cos \left( {\frac{2}{x}} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} = \mathop {\lim }\limits_{x \to + \infty } {e^{{x^2}\ln \cos \left( {\frac{2}{x}} \right)}} = {e^{\mathop {\lim }\limits_{x \to + \infty } {x^2}\ln \cos \left( {\frac{2}{x}} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to + \infty } {x^2}\ln \cos \left( {\frac{2}{x}} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln \cos \left( {\frac{2}{x}} \right)}}{{\frac{1}{{{x^2}}}}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{d}{{dx}}\left[ {\ln \cos \left( {\frac{2}{x}} \right)} \right]}}{{\frac{d}{{dx}}\left[ {\frac{1}{{{x^2}}}} \right]}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - \frac{2}{{{x^2}}}\tan \left( {\frac{2}{x}} \right)}}{{ - \frac{2}{{{x^3}}}}} \cr
& L = \mathop {\lim }\limits_{x \to + \infty } \frac{{\tan \left( {\frac{2}{x}} \right)}}{{\frac{1}{x}}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to + \infty } \frac{{\tan \left( {\frac{2}{x}} \right)}}{{\frac{1}{x}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{2}{{{x^2}}}{{\sec }^2}\left( {\frac{2}{x}} \right)}}{{ - \frac{1}{{{x^2}}}}} = - \mathop {\lim }\limits_{x \to + \infty } 2{\sec ^2}\left( {\frac{2}{x}} \right) = - 2 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} = \mathop {\lim }\limits_{x \to + \infty } {e^{{x^2}\ln \cos \left( {\frac{2}{x}} \right)}} = {e^{\mathop {\lim }\limits_{x \to + \infty } {x^2}\ln \cos \left( {\frac{2}{x}} \right)}} = {e^{ - 2}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} = \frac{1}{{{e^2}}} \cr} $$
