Question

If $p=\underset{x\to \infty }{lim}(\sin \sqrt{x^2+1}-\sin |x|)$ and $q=\underset{x\to -\infty }{lim}\left[\sin \left(\cos {\left(\sqrt{|x|-x^3}\right)}^{-1}\right)\right]$
(Where [.] denotes greatest integer function), then which of the following is/are correct:

• A. $p=0$
• B. $q=0$
• C. $p=1$
• D. $q=-1$

Answer 1

Answer: (A), (B)

$q=0$

$p=\underset{x\to \infty }{lim}(\sin \sqrt{x^2+1}-\sin |x|)$
$=\underset{x\to \infty }{lim}\left(\sin \sqrt{x^2+1}\right)-\sin x$
$=\underset{x\to \infty }{lim}2\cos \left(\frac{\sqrt{x^2+1}+x}{2}\right)\sin \left(\frac{\sqrt{x^2+1}-x}{2}\right)$
$=\underset{x\to \infty }{lim}2\cos \left(\frac{\sqrt{x^2+1}+x}{2}\right)\sin \left(\frac{1}{2\sqrt{x^2+1}+x}\right)=0$
$\Rightarrow [\because \cos \left(\frac{\sqrt{x^2+1}+x}{2}\right)\text{is finite}\forall x\in R]$
$\therefore p=0$
Now, $q=\underset{x\to -\infty }{lim}\left[\sin \left(\cos {\left(\sqrt{|x|-x^3}\right)}^{-1}\right)\right]$
$\Rightarrow q=\underset{x\to -\infty }{lim}\left[\sin \left(\cos \left(\frac{1}{\sqrt{-(x+x^3)}}\right)\right)\right].$
$q=\left[\sin \right(\cos 0\left)\right]=\left[\sin 1\right]=0$