The value of $$\lim_{x \to 0} \frac{1}{x}\left[\tan^{-1}\left(\frac{x+1}{2x+1}\right) - \frac{\pi}{4}\right]$$ is :

$$L=\lim_{x\to0}\frac{1}{x}\left[\tan^{-1}\left(\frac{x+1}{2x+1}\right)-\frac{\pi}{4}\right]$$
$$\tan^{-1}a-\tan^{-1}b$$

$$\tan^{-1}\left(\frac{a-b}{1+ab}\right)$$

with (b=1):
$$L=\lim_{x\to0}\frac{1}{x}$$
$$\tan^{-1}\left(\frac{\frac{x+1}{2x+1}-1}{1+\frac{x+1}{2x+1}}\right)$$

Simplify:$$\frac{\frac{x+1-(2x+1)}{2x+1}}{\frac{3x+2}{2x+1}}$$

$$\frac{-x}{3x+2}$$
$$L=\lim_{x\to0}\frac{1}{x}\tan^{-1}\left(\frac{-x}{3x+2}\right)$$
Using $$(\tan^{-1}t\sim t)as(t\to0),$$

$$L=\lim_{x\to0}\frac{1}{x}\cdot\frac{-x}{3x+2}= -\frac{1}{2}$$