VMO

a. We will prove $x_n\ge n^2$ for all positive integers by induction. The statement is obvious in case $n=1$. Assume that $x_n\ge n^2$, hence $$x_{n+1}>x_n+3\sqrt{x_n}\ge x_n+3n\ge (n+1{)}^2.$$ Therefore, the statement is proved. Thus $$0<\frac{n}{x_n}\ge \frac{1}{n}.$$ By the Squeeze theorem, one can get ${\lim }_{n\to +\infty }\frac{n}{x_n}=0$.

b. Let $x_n=y_n^2$ and the given formula can be rewritten as $$(y_{n+1}-y_n)(y_{n+1}+y_n)=3y_n+\frac{n}{y_n}$$ then combine with the formula of $y_{n+1}$ calculated by $y_n$, the above equality obtains that $$\begin{array}{rl}{y}_{n+1}-y_n& =\frac{3y_n+\frac{n}{y_n}}{\sqrt{y_n^2+3y_n+\frac{n}{y_n}}+y_n}\\ & =\frac{3+\frac{n}{y_n^2}}{\sqrt{1+\frac{3}{y_n}+\frac{n}{y_n^3}}+1}\end{array}$$ From part a), one could deduce that $$\underset{n\to +\infty }{\lim }\frac{1}{y_n}=\underset{n\to +\infty }{\lim }\frac{n}{y_n^3}=0$$ hence $$\underset{n\to +\infty }{\lim }(y_{n+1}-y_n)=\frac{3}{2}.$$ By Cesàro mean theorem, $$\underset{n\to +\infty }{\lim }\frac{y_n}{n}=\frac{3}{2}\Rightarrow \underset{n\to +\infty }{\lim }\frac{n^2}{x_n}=\frac{4}{9}$$ and it is the result.