Using definition of limit of Sequence show that, If x>0 then limit of Sequence (Arctan nx) is π/2

Question

Question: I need to show by "using definition of limit of Sequence" that Sequence $(\arctan nx)$ converges to $π/2$ if $x>0$.

My attempt: i need to show for any $\epsilon >0$, and for each $x>0$, we have $|\arctan nx -π/2|<\epsilon$ for sufficiently large $n$. How do i proceed further, unable to solve it. Please help.

Answer 1

Hint: following up on K.defaoite's reply, consider any integer $$N > \frac{\tan(\pi/2 - \varepsilon)}{x}$$ and use that as the lower bound for sufficiently large $n$.

Edit: I've put a full solution in the spoiler box below.

Let $\varepsilon > 0$. We wish to show that there exists a sufficiently large integer $N$ for which $\lvert\arctan{nx} - \pi/2\rvert < \varepsilon$ for all $n \geq N$. Define $\varepsilon_0 = \min(\varepsilon, \pi/4)$, and take $N$ to be any integer such that $$N > \frac{\tan(\pi/2 - \varepsilon_0)}{x}.$$ Then for all $n \geq N$, \begin{align*}\lvert\arctan{nx} - \pi/2\rvert &= \pi/2 - \arctan(nx)\\ &\leq \pi/2 - \arctan(Nx) \\ &< \pi/2 - \arctan\bigg(\frac{\tan(\pi/2 - \varepsilon_0)}{x} \cdot x \bigg) \\ &= \pi/2 - (\pi/2 - \varepsilon_0) \\ &\leq \varepsilon, \end{align*} as desired. Note that the first equality holds because $\pi/2 > \arctan{x}$ for all $x \in \mathbb{R}$, and the inequalities in lines 2 and 3 hold because $\arctan$ is increasing on $\mathbb{R}$.