Question

The limit $\underset{x\to \infty }{\lim }x\left[{\tan }^{-1}\left(\frac{x+1}{x+2}\right)-{\tan }^{-1}\left(\frac{x}{x+2}\right)\right]$ is equal to

• A. $2$
• B. $\frac{1}{2}$
• C. $-\frac{1}{3}$
• D. None of these

Answer 1

Answer: (B)

$\underset{x\to \infty }{\lim }x\left[{\tan }^{-1}\left(\frac{x+1}{x+2}\right)-{\tan }^{-1}\left(\frac{x}{x+2}\right)\right]$
$=\underset{x\to \infty }{\lim }x{\tan }^{-1}\left(\frac{\frac{x+1}{x+2}-\frac{x}{x+2}}{1+\frac{x+1}{x+2}\cdot \frac{x}{x+2}}\right)$
$=\underset{x\to \infty }{\lim }x{\tan }^{-1}\left(\frac{x+2}{2x^2+5x+4}\right)$
$=\underset{x\to \infty }{\lim }x\left(\frac{{\tan }^{-1}\left(\frac{x+2}{2x^2+5x+4}\right)}{\frac{x+2}{2x^2+5x+4}}\right)\times \frac{x(x+2)}{2x^2+5x+4}=1\times \frac{1}{2}=\frac{1}{2}$