<s>[INST] Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$ 
Evaluate 
  $$\lim_{ x\to \infty} \left(  \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$

I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left(  \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
 [/INST] $$
\begin{aligned}
\lim _{x\to \infty }\left(\arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi \:}{4}\right)x
& = \lim _{t\to 0}\left(\frac{\arctan \left(\frac{t+1}{4t+1}\right)-\frac{\pi }{4}}{t}\right) \\
& = \lim _{t\to 0}\left(\frac{\left(\frac{\pi \:}{4}-\frac{3}{2}t+o\left(t\right)\right)-\frac{\pi \:}{4}}{t}\right)\\
& = \color{red}{-\frac{3}{2}}
\end{aligned}
$$
Solved with Taylor expansion:
$$\arctan \left(\frac{t+1}{4t+1}\right) = \frac{\pi }{4}-\frac{3t}{2}+\frac{15t^2}{4}-\frac{33t^3}{4}+15t^4+\ldots \:$$
</s>