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,

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

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

