Answer
$1$
Work Step by Step
To find the limit
$$
\lim _{x \rightarrow \infty}x^{1/x^2}
$$
we have to find
$$\ln \lim _{x \rightarrow \infty}x^{1/x^2}=\lim _{x \rightarrow \infty}\ln x^{1/x^2}\\
=\lim _{x \rightarrow \infty}\frac{\ln x}{x^2}=\frac{\infty}{\infty}$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow \infty}\frac{\ln x}{x^2}=\lim _{x \rightarrow \infty}\frac{1}{2x^2}=0.
$$
Now, $$\ln \lim _{x \rightarrow \infty}x^{1/x^2}=0\Longrightarrow \lim _{x \rightarrow \infty}x^{1/x^2}=e^0=1. $$
