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

Yes Possible.
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
$$L=\lim_{y \to 0}\frac{\left(  tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
$$L=\lim_{y \to 0}\frac{\left(  tan^{-1}\left(\frac{1+y}{1+4y}\right)-\tan^{-1}{1}\right)}{y}$$
using the following formula, we can acheive, $$tan^{-1}A-tan^{-1}B = \frac{A-B}{1+AB}$$ we get
$$L=\lim_{y \to 0} \frac{1}{y} \times \frac{-3y}{(2+5y)}$$
$$L=\lim_{y \to 0} \frac{-3}{(2+5y)}=\frac{-3}{2}$$
