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

Let 
$$y=\arctan\left(\frac{1+x}{4+x}\right).$$
Then 
$$x=\frac{1}{2}\left[3\left(\frac{1+\tan y}{1-\tan y}\right)-5\right].$$
As $x \to \infty, y \to \frac{\pi}{4}.$
So your limit is
\begin{align*}
\lim_{ x\to \infty} \left(  \arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x & = \lim_{y \to \pi/4}\left(y-\frac{\pi}{4}\right)\frac{1}{2}\left[3\left(\frac{1+\tan y}{1-\tan y}\right)-5\right]\\
& = \lim_{y \to \pi/4}\left(y-\frac{\pi}{4}\right)\frac{1}{2}\left[3\left(\frac{-1}{\tan\left(y-\frac{\pi}{4}\right)}\right)-5\right]
\end{align*}
Now with $\theta=y-\pi/4$, use the limit $\lim_{\theta \to 0}\frac{\tan \theta}{\theta}=1$ to get the answer as $-3/2$. 
