Basic limit ,arctan

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{n \to \infty} \left( n \tan^{-1} 3n - n \tan^{-1} n \right) \\ & = \lim_{n \to \infty} n \tan^{-1} \left(\frac {3n-n}{1+3n^2} \right) \\ & = \lim_{n \to \infty} n \color{#3D99F6} \tan^{-1} \left(\frac {2n}{1+3n^2} \right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{n \to \infty} n \color{#3D99F6} \left(\frac {2n}{1+3n^2} - \frac 13 \left(\frac {2n}{1+3n^2}\right)^3 + \frac 15 \left(\frac {2n}{1+3n^2}\right)^5 - \cdots \right) \\ & = \lim_{n \to \infty} \left(\frac {2n^2}{1+3n^2} - \frac {2^3n^4}{3(1+3n^2)^3} + \frac {2^5n^6}{5(1+3n^2)^5} - \cdots \right) \\ & = \lim_{n \to \infty} \left(\frac {2}{n^{-2}+3} - \frac {2^3n^{-2}}{3(n^{-2}+3)^3} + \frac {2^5n^{-4}}{5(n^{-2}+3)^5} - \cdots \right) \\ & = \frac 23 \approx \boxed{0.667} \end{aligned}$

$\lim\limits_{n \to \infty}\left\{n\arctan{(3n)}-n\arctan{(n)}\right\}=\lim\limits_{n \to \infty}\frac{\arctan{(3n)}-\arctan{(n)}}{\frac{1}{n}}="\frac{0}{0}"$

You can continue solving using $L'Hôpital's$ rule .