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
 A: 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
$$L=\lim_{y \to 0} \frac{1}{y} \times \frac{-3y}{(2+5y)}$$
$$L=\lim_{y \to 0} \frac{-3}{(2+5y)}=\frac{-3}{2}$$
A: HINT:
Set $1/x=h$ to get $$F=\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1}h$$
Now $\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1=\tan^{-1}\left(\dfrac{\dfrac{h+1}{4h+1}-1}{1+\dfrac{h+1}{4h+1}}\right)=\tan^{-1}\dfrac{-3h}{5h+2}$
$$\implies F=\lim_{h\to0^+}\dfrac{\tan^{-1}\left(\dfrac{-3h}{5h+2}\right)}{\left(\dfrac{-3h}{5h+2}\right)}\cdot\lim_{h\to0^+}\dfrac{-3}{5h+2}$$
A: Let $\arctan\frac{1+x}{4+x}=\frac{\pi}{4}+y$.
Hence, $y\rightarrow0$ and $$\frac{1+x}{4+x}=\tan\left(\frac{\pi}{4}+y\right)$$
or $$x=\frac{1-4\tan\left(\frac{\pi}{4}+y\right)}{\tan\left(\frac{\pi}{4}+y\right)-1}$$ and we need to calculate
$$\lim_{y\rightarrow0}\frac{y\left(1-4\tan\left(\frac{\pi}{4}+y\right)\right)}{\tan\left(\frac{\pi}{4}+y\right)-1}$$ or
$$-\frac{3}{\sqrt2}\lim_{y\rightarrow0}\frac{y}{\sin\left(\frac{\pi}{4}+y\right)-\cos\left(\frac{\pi}{4}+y\right)}$$ or
$$-\frac{3}{\sqrt2}\lim_{y\rightarrow0}\frac{y}{\frac{2}{\sqrt2}\sin{y}},$$
which is $-\frac{3}{2}$.
A: 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.
A: 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 $-3/2$. 
A: 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 $-3/2$.
A: 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,
$$
\begin{align}
\lim_{x\to\infty}\left(\tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac\pi4\right)x
&=\lim_{x\to\infty}\tan^{-1}\left(\frac{-3}{5+2x}\right)x\\
&=\lim_{x\to\infty}\frac{\tan^{-1}\left(\frac{-3}{5+2x}\right)}{\frac{-3}{5+2x}}\lim_{x\to\infty}\frac{-3x}{5+2x}\\
&=-\frac32
\end{align}
$$
A: $$
\begin{aligned}
\lim _{x\to \infty }\left(\arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi \:}{4}\right)x
& = \lim _{t\to 0}\left(\frac{\arctan \left(\frac{t+1}{4t+1}\right)-\frac{\pi }{4}}{t}\right) \\
& = \lim _{t\to 0}\left(\frac{\left(\frac{\pi \:}{4}-\frac{3}{2}t+o\left(t\right)\right)-\frac{\pi \:}{4}}{t}\right)\\
& = \color{red}{-\frac{3}{2}}
\end{aligned}
$$
Solved with Taylor expansion:
$$\arctan \left(\frac{t+1}{4t+1}\right) = \frac{\pi }{4}-\frac{3t}{2}+\frac{15t^2}{4}-\frac{33t^3}{4}+15t^4+\ldots \:$$
