Question

$\underset{x\to 0}{lim}\frac{\sin \left[{\pi }^2\right]x-\sin [-{\pi }^2]x}{{\tan }^2\sqrt{x}}$ is equal to ( where [.] denotes greatest integer function )

• A. 0
• B. 18
• C. 19
• D. does not exist

Answer 1

Answer: (C)

19

Given, $\underset{x\to 0}{lim}\frac{\sin \left[{\pi }^2\right]x-\sin [-{\pi }^2]x}{{\tan }^2\sqrt{x}}$
$(\because {\pi }^2\simeq 9.85)$
= $\underset{x\to 0}{lim}\frac{\sin 9x+\sin 10x}{{\left(\sqrt{x}\right)}^2}=9+10=19$