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

Well, we can start as you, by setting $y=\frac{1}{x}$. Now, our limits transforms to:
$$L=\lim_{y\to0}\frac{\tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}}{y}$$
Now, let $f:\mathbb{R}\to\mathbb{R}$ with 
$$f(x)=\tan\left(\frac{1+x}{1+4x}\right)$$
Note that $f(0)=\tan^{-1}(1)=\frac{\pi}{4}$. So, we have:
$$L=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)$$
since $f$ is differentiable.
Now, $f$'s formula is:
$$f(x)=\int_0^\frac{1+x}{1+4x}\frac{1}{1+t^2}dt$$
So, we have:
$$f'(x)=\frac{1}{1+\left(\frac{1+x}{1+4x}\right)^2}\left(\frac{1+x}{1+4x}\right)'=\frac{(1+4x)^2}{1+(1+x)^2}\frac{-3}{(1+4x)^2}=\frac{-3(1+4x)^2}{(1+(1+x)^2)(1+4x)^2}$$
So
$$L=f'(0)=-\frac{3}{2}$$
As you have already proved.
