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