Question

The value of$\underset{x\to \infty }{lim}x[{\tan }^{-1}\frac{x+1}{x+2}-{\cot }^{-1}\frac{x+2}{x}]$ is:

• A. $-1$
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (C)

$\frac{1}{2}$

$\underset{x\to \infty }{lim}x[{\tan }^{-1}\frac{x+1}{x+2}-{\cot }^{-1}\frac{x+2}{x}]$

$=\underset{x\to \infty }{lim}x[{\tan }^{-1}\frac{x+1}{x+2}-{\tan }^{-1}\frac{x}{x+2}]$

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

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

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

$=\underset{x\to \infty }{lim}\frac{x^2+2x}{2x^2+5x+4}$

$=\underset{x\to \infty }{lim}\frac{x^2(1+\frac{2}{x})}{x^2(2+\frac{5}{x}+\frac{4}{x^2})}$

$=\frac{1}{2}$