Question

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

• A. $-20+\tan 20$
• B. $20+\tan 20$
• C. $20$
• D. $\tan 20$

Answer 1

The correct option is A $-20+\tan 20$

We know, $-19<-2{\pi }^2<-20\therefore [-2{\pi }^2]=-20$
Thus,
$\underset{x\to 0}{lim}\frac{\tan \left(\right[-2{\pi }^2\left]x^2\right)-x^2\tan \left(\right[-2{\pi }^2\left]\right)}{{\sin }^{2}x}$
$=\underset{x\to 0}{lim}\frac{\tan (-20x^2)-x^2\tan (-20)}{{\sin }^{2}x}$
$=\underset{x\to 0}{lim}\frac{x^2\tan \left(20\right)-\tan \left(20x^2\right)}{{\sin }^{2}x}$
$=\underset{x\to 0}{lim}\frac{\tan \left(20\right)-20\cdot \left(\frac{\tan \left(20x^2\right)}{20x^2}\right)}{{\left(\frac{\sin x}{x}\right)}^2}$
$=\tan \left(20\right)-20$