Calculate this limit : $\lim_{x\rightarrow +\infty}\left[x\left(4\arctan\left(\frac{x+1}{x}\right)-\pi\right)\right]$ Calculate the limit
$$\lim_{x\rightarrow +\infty}\left[x\left(4\arctan\left(\frac{x+1}{x}\right)-\pi\right)\right]$$
Neither L'Hospital's rule nor Taylor expansions are allowed,
 A: Herein, we present an approach that relies on only (1) a set of inequalities for the arctangent function, obtained using only elementary geometry, and (2) the squeeze theorem.  To that end, we begin with the following primer.

PRIMER:
I showed in THIS ANSWER, using only elementary inequalities from geometry, that the arctangent function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{\sqrt{1+x^2}} \le \arctan(x) \le x} \tag 1$$
for $x\ge 0$.


Note that we can write
$$\arctan\left(\frac{x+1}{x}\right)=\pi/4+\arctan\left(\frac{1}{2x+1}\right)$$
Therefore, we see that
$$4\arctan\left(\frac{x+1}{x}\right)-\pi= \arctan\left(\frac4{2x+1}\right) \tag 2$$
Combining $(1)$ and $(2)$ reveals
$$\frac{\frac{4x}{2x+1}}{\sqrt{1+\left(\frac{1}{2x+1}\right)^2}} \le x\,\left(4\arctan\left(\frac{x+1}{x}\right)-\pi\right) \le \frac{4x}{2x+1}$$
whereupon application of the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}x\,\left(4\arctan\left(\frac{x+1}{x}\right)-\pi\right)=2}$$

A: Let $4\arctan\left(1+\frac1x\right)-\pi=4y,$ then we have $x=\dfrac{1}{\tan (\pi/4+y)-1}=\dfrac{1-\tan y}{2\tan y}.$
 Now the required limit equals to $$\lim_{y\to 0}2y\left(\dfrac{1-\tan y}{\tan y}\right)=2$$ as $\lim_{y\to 0}\dfrac{\tan y}{y}=1.$
A: A solution without L'Hospital and Taylor which uses only the geometric series and basic differential calculus
Consider the function
$$
f(x)=4\arctan\left(\frac{x+1}{x}\right),\quad f(\infty)=4\arctan(1)=\pi
$$
Now
$$
f'(x)=-\frac{4}{2 x (x+1)+1}
$$
By a simple application of the geometric series we see that
$$
f'(x)\sim_{\infty}-\frac{2}{x^2}
$$
or
$$
f(x)\sim_{\infty}\frac{2}{x}+C
$$
where $C$ is fixed by the conditon $f(\infty)=\pi$ so
$$
f(x)\sim_{\infty}\frac{2}{x}+\pi
$$
so we have

$$
\lim_{x\rightarrow\infty}[x(f(x)-\pi)]=2
$$

A: It is enough to exploit some trigonometric manipulation. Since $\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$ we have:
$$ \arctan\left(1+\frac{1}{x}\right) = \frac{\pi}{4}+\arctan\frac{1}{2x+1} \tag{1}$$
and our limit equals (since $\lim_{z\to 0}\frac{\arctan z}{z}=1$)
$$ \lim_{x\to +\infty}4x\arctan\frac{1}{2x+1} =\lim_{x\to +\infty}\frac{4x}{2x+1}=\color{red}{\large 2}.\tag{2}$$
