Inverse Trigonometric Functions Question 52
Question: The limit $ \underset{x\to \infty }{\mathop{\lim }}x[ {{\tan }^{-1}}( \frac{x+1}{x+2} )-{{\tan }^{-1}}( \frac{x}{x+2} ) ] $ is equal to
Options:
A) 2
B) $ \frac{1}{2} $
C) $ -\frac{1}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
$ \underset{x\to \infty }{\mathop{\lim }}x[ {{\tan }^{-1}}( \frac{x+1}{x+2} )-{{\tan }^{-1}}( \frac{x}{x+2} ) ] $
$ =\underset{x\to \infty }{\mathop{Lim}}x{{\tan }^{-1}}( \frac{\frac{x+1}{x+2}-\frac{x}{x+2}}{1+\frac{x+1}{x+2}.\frac{x}{x+2}} ) $
$ =\underset{x\to \infty }{\mathop{Lim}}x{{\tan }^{-1}}( \frac{x+2}{2x^{2}+5x+4} ) $
$ =\underset{x\to \infty }{\mathop{Lim}}x( \frac{{{\tan }^{-1}}( \frac{x+2}{2x^{2}+5x+4} )}{\frac{x+2}{2x^{2}+5x+4}} )\times \frac{x(x+2)}{2x^{2}+5x+4} $
$ =1\times \frac{1}{2}=\frac{1}{2} $
BETA
