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

Using the formula for the tangent of a sum,
$$
\tan\left(x-\frac\pi4\right)=\frac{\tan(x)-1}{\tan(x)+1}
$$
we get
$$
\tan^{-1}(x)-\frac\pi4=\tan^{-1}\left(\frac{x-1}{x+1}\right)
$$
Thus,
$$
\begin{align}
\lim_{x\to\infty}\left(\tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac\pi4\right)x
&=\lim_{x\to\infty}\tan^{-1}\left(\frac{-3}{5+2x}\right)x\\
&=\lim_{x\to\infty}\frac{\tan^{-1}\left(\frac{-3}{5+2x}\right)}{\frac{-3}{5+2x}}\lim_{x\to\infty}\frac{-3x}{5+2x}\\
&=-\frac32
\end{align}
$$
