Question

If $[\cdot ]$ denotes the greatest integer function, then $\underset{x\to 0}{\lim }\frac{\tan \left(\left[-2{\pi }^2\right]x^2\right)-x^2\tan \left[-2{\pi }^2\right]}{{\sin }^{2}x}$ is equal to

• A. $-20+\tan 20$
• B. $20+\tan 20$
• C. $20$
• D. None of these

Answer 1

Answer: (A)

We have, $\left[-2{\pi }^2\right]=-20$
$\Rightarrow \underset{x\to 0}{\lim }\frac{\tan \left(\left[-2{\pi }^2\right]x^2\right)-x^2\tan \left[-2{\pi }^2\right]}{{\sin }^{2}x}$
$=\underset{x\to 0}{\lim }\frac{\tan \left(-20x^2\right)-x^2\tan (-20)}{{\sin }^{2}x}$
$=\underset{x\to 0}{\lim }-\frac{\tan 20x^2}{20x^2}\times 20\times \frac{x^2}{{\sin }^{2}x}+{\left(\frac{x}{\sin x}\right)}^2\tan 20$
$=-20+\tan 20$