Inverse fun

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to \infty} \frac {(x+5)\tan^{-1}(x+5)-(x+1)\tan^{-1}(x+1)}{\sqrt{x^2+1}+\tan^{-1} x-x} \\ & = \lim_{x \to \infty} \frac {x\left(\tan^{-1}(x+5)-\tan^{-1}(x+1)\right)+5\tan^{-1}(x+5)-\tan^{-1}(x+1)}{\frac {\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\sqrt{x^2+1}+x}+\tan^{-1} x} \\ & = \lim_{x \to \infty} \frac {x\left(\tan^{-1}(x+5)-\tan^{-1}(x+1)\right)+5\tan^{-1}(x+5)-\tan^{-1}(x+1)}{\frac 1{\sqrt{x^2+1}+x}+\tan^{-1} x} \\ & = \lim_{x \to \infty} \frac {{\color{#3D99F6}\frac {\tan^{-1}(x+5)-\tan^{-1}(x+1)}{1/x}}+5\tan^{-1}(x+5)-\tan^{-1}(x+1)}{\frac 1{\sqrt{x^2+1}+x}+\tan^{-1} x} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to \infty} \frac {{\color{#3D99F6}\frac {1/(1+(x+5)^2)-1/(1+(x+1)^2)}{-1/x^2}}+5\tan^{-1}(x+5)-\tan^{-1}(x+1)}{\frac 1{\sqrt{x^2+1}+x}+\tan^{-1} x} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t }x. \\ & = \lim_{x \to \infty} \frac {{\color{#3D99F6}\frac {x^2}{1+(x+1)^2}-\frac {x^2}{1+(x+5)^2}}+5\tan^{-1}(x+5)-\tan^{-1}(x+1)}{\frac 1{\sqrt{x^2+1}+x}+\tan^{-1} x} & \small \color{#3D99F6} \text{Divide up and down by }x^2. \\ & = \lim_{x \to \infty} \frac {{\color{#3D99F6}\frac 1{1/x^2+(x+1)^2/x^2}-\frac 1{1/x^2+(x+5)^2/x^2}}+5\tan^{-1}(x+5)-\tan^{-1}(x+1)}{\frac 1{\sqrt{x^2+1}+x}+\tan^{-1} x} \\ & = \frac {{\color{#3D99F6}1-1}+\frac {5\pi} 2-\frac \pi 2}{0+\frac \pi 2} \\ & = \boxed{4} \end{aligned}$