<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] It would be much simpler to proceed directly without any change of variables. Note that the expression under limit can be written as $$x\arctan\dfrac{\dfrac{1+x}{4+x}-1}{\dfrac{1+x}{4+x}+1}$$ which is same as $$-\frac{3x}{5+2x}\cdot\dfrac{\arctan\dfrac{3}{5+2x}}{\dfrac{3}{5+2x}}$$ and first factor tends to $-3/2$ and second factor tends to $1$ so that the desired limit is $-3/2$.
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