For any positive integer n , define f _ n : (0, ∞) → R as f _ n (x)...

{f}_{n}\left(x\right) = \mathop{∑}_{j = 1}^{n}\tan^{- 1}\left(\dfrac{\left(x + j\right) - \left(x + j - 1\right)}{1 + \left(x + j\right)\left(x + j - 1\right)}\right)${f}_{n}\left(x\right) = \mathop{∑}_{j = 1}^{n}\left[\tan^{- 1}\left(x + j\right) - \tan^{- 1}\left(x + j - 1\right)\right]$
${f}_{n}\left(x\right) = \tan^{- 1}\left(x + n\right) - \tan^{- 1}x$
$∴$ $\tan \left({f}_{n}\left(x\right)\right) = \tan \left[\tan^{- 1}\left(x + n\right) - \tan^{- 1}x\right]$

$\tan \left({f}_{n}\left(x\right)\right) = \tan \left[\tan^{- 1}\dfrac{\left(x + n\right) - x}{1 + x\left(x + n\right)}\right]$
$\tan \left({f}_{n}\left(x\right)\right) = \dfrac{n}{1 + {x}^{2} + nx}$,

so that $\lim_{x \to \infty}\tan \left({f}_{n}\left(x\right)\right) = 0$
$∴$ $\sec^{2}\left({f}_{n}\left(x\right)\right) = 1 + \tan^{2}\left({f}_{n}\left(x\right)\right)$
$\sec^{2}\left({f}_{n}\left(x\right)\right) = 1 + {\left(\dfrac{n}{1 + {x}^{2} + nx}\right)}^{2}$
$\lim_{x \to \infty}\sec^{2}\left({f}_{n}\left(x\right)\right) = \lim_{x \to \infty}1 + {\left(\dfrac{n}{1 + {x}^{2} + nx}\right)}^{2} = 1$

Note that since $0$ is not in domain so first two options are wrong