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

Correct option is B

Given:
​$$\lim_{x \to \infty} x \left[ \tan^{-1}\!\left( \frac{x+1}{x+2} \right) - \frac{\pi}{4} \right]$$​​
Formula used:
For small h, $$\tan^{-1}(1 + h) = \frac{\pi}{4} + \frac{h}{2} + o(h)$$​​
Solution:
​$$\frac{x+1}{x+2} = 1 - \frac{1}{x+2}$$​​
So,
​$$\tan^{-1}\!\left( \frac{x+1}{x+2} \right)= \tan^{-1}\!\left( 1 - \frac{1}{x+2} \right)$$​​
Using the expansion:
​$$\tan^{-1}\!\left( 1 - \frac{1}{x+2} \right)= \frac{\pi}{4} - \frac{1}{2(x+2)} + o\!\left( \frac{1}{x} \right)$$​​
Hence,
​$$x \left[ \tan^{-1}\!\left( \frac{x+1}{x+2} \right) - \frac{\pi}{4} \right]= x \left( -\frac{1}{2(x+2)} \right)\\= -\frac{x}{2(x+2)}$$​​
Taking limit as $$x \to \infty:$$​​
​$$\lim_{x \to \infty} \left( -\frac{x}{2(x+2)} \right)= -\frac{1}{2}$$​​
The correct answer is (b) $$-\frac{1}{2}.$$​​