Answer
$ e^{-3/2}.$
Work Step by Step
To find the limit
$$
\lim _{x \rightarrow 0}(\cos x)^{3/x^2}
$$
we have to find
$$\ln \lim _{x \rightarrow 0}(\cos x)^{3/x^2}
=\lim _{x \rightarrow 0}\ln(\cos x)^{3/x^2}=\lim _{x \rightarrow 0}\frac{3\ln \cos x}{x^2}
=\frac{\infty}{\infty}$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0}\frac{3\ln \cos x}{x^2}=\lim _{x \rightarrow 0}\frac{-3\sin x/ \cos x}{2x}=\frac{0}{0}
$$
Which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow 0}\frac{-3\sin x/ \cos x}{2x}=\lim _{x \rightarrow 0}\frac{-3\sin x}{2x\cos x}=\lim _{x \rightarrow 0}\frac{-3\cos x}{2\cos x-2x\sin x}=-\frac{3}{2}.$$
Now, $$\ln \lim _{x \rightarrow 0}(\cos x)^{3/x^2}=-\frac{3}{2}.\Longrightarrow \lim _{x \rightarrow 0}(\cos x)^{3/x^2}=e^{-3/2}. $$
