Question

$\underset{x\to \infty }{lim}\left\{\right(x+5){\tan }^{-1}(x+5)-(x+1\left){\tan }^{-1}(x+1)\right\}$ is equal to

• A. $\pi$
• B. $2\pi$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

The correct option is B $2\pi$

$\text{Given Limit is,}\underset{x\to \infty }{lim}(x+1)\left[{\tan }^{-1}\right(x+5)-(x+1\left)\right]+4{\tan }^{-1}(x+5)$
$=\underset{x\to \infty }{lim}\left[\right(x+1\left){\tan }^{-1}\frac{4}{1+(x+1)(x+5)}+4{\tan }^{-1}(x+5)\right]$
$=\underset{x\to \infty }{lim}\left[\right(x+1\left){\tan }^{-1}\frac{\frac{4}{x^2+6x+6}}{\left(\frac{4}{x^2+6x+6}\right)}\times \frac{4}{x^2+6x+6}+4{\tan }^{-1}(x+5)\right]$
$=0+4\times \frac{\pi }{2}=2\pi$