Answer
By Squeeze Theorem:
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{\arctan \left( {{n^2}} \right)}}{{\sqrt n }} = 0$
Work Step by Step
From the information prior to Exercise 31 in Section 11.3, we know that for $n$ sufficiently large, the following inequality is valid:
$\ln n \le {n^a}$, ${\ \ \ \ \ }$ for all $a \gt 0$
Let $a = \dfrac{1}{2}$. So, $\ln n \le \sqrt n $.
Therefore,
$\dfrac{1}{{\sqrt n }} \le \dfrac{1}{{\ln n}}$
For $n \gt 0$, we have $\arctan \left( {{n^2}} \right) \gt 0$ and increasing. So,
$0 \le \dfrac{{\arctan \left( {{n^2}} \right)}}{{\sqrt n }} \le \dfrac{{\arctan \left( {{n^2}} \right)}}{{\ln n}}$
Take the limit on all sides:
$\mathop {\lim }\limits_{n \to \infty } 0 \le \mathop {\lim }\limits_{n \to \infty } \dfrac{{\arctan \left( {{n^2}} \right)}}{{\sqrt n }} \le \mathop {\lim }\limits_{n \to \infty } \dfrac{{\arctan \left( {{n^2}} \right)}}{{\ln n}}$
As $n \to \infty $, $\arctan \left( {{n^2}} \right) \to \dfrac{\pi }{2}$ and $\ln n \to \infty $. Therefore, $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\arctan \left( {{n^2}} \right)}}{{\ln n}} = 0$.
Thus,
$0 \le \mathop {\lim }\limits_{n \to \infty } \dfrac{{\arctan \left( {{n^2}} \right)}}{{\sqrt n }} \le 0$
By Squeeze Theorem, $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\arctan \left( {{n^2}} \right)}}{{\sqrt n }} = 0$.
