Answer
$x^2\ln x$ grows slower than $x^3$.
Work Step by Step
We will find the limit $\lim_{x\to\infty}\frac{x^2\ln x}{x^3}.$
1) If it is equal to zero then $x^2\ln x$ grows slower than $x^3$;
2) If it is equal to $\infty$ then $x^2\ln x$ grows faster than $x^3$;
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":
$$\lim_{x\to\infty}\frac{x^2\ln x}{x^3}=\lim_{x\to\infty}\frac{\ln x}{x}=\left[\frac{\ln\infty}{\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln x)'}{(x)'}=\lim_{x\to\infty}\frac{\frac{1}{x}}{1}=\lim_{x\to\infty}\frac{1}{x}=\left[\frac{1}{\infty}\right]=0,$$
and thus $x^2\ln x$ grows slower than $x^3$.
