Answer
$x^x$ grows faster than $(x/2)^x$.
Work Step by Step
We will find the limit $\lim_{x\to\infty}\frac{x^x}{(x/2)^x}.$
1) If it is equal to zero then $x^x$ grows slower than $(x/2)^x$;
2) If it is equal to $\infty$ then $x^x$ grows faster than $(x/2)^x$;
3) If it is equal to some non zero constant then their growth rates are comparable.
"LR" will stand for "Apply L'Hopital's rule":
$$\lim_{x\to\infty}\frac{x^x}{\left(\frac{x}{2}\right)^x}=\lim_{x\to\infty}\left(\frac{x}{\frac{x}{2}}\right)^x=\lim_{x\to\infty}\left(2\right)^x=\left[2^\infty\right]=\infty,$$
and thus $x^x$ grows faster than $(x/2)^x$.
