Question

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

• A. $\tan 10-10$
• B. ${\tan }^{2}10-20$
• C. $\tan 10$
• D. $0$

Answer 1

The correct option is A $\tan 10-10$

We know $-10<{\pi }^2<-10\Rightarrow [-{\pi }^2]=-10$

$\therefore Limit=\underset{x\to 0}{lim}\frac{\tan [-{\pi }^2]x^2-\left\{\tan \right[-{\pi }^2\left]\right\}.x^2}{{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\frac{\tan (-10)x^2-\left\{\tan \right(-10\left)\right\}.x^2}{{\sin }^{2}x}=\underset{x\to 0}{lim}\frac{\tan \left(10\right)\cdot x^2-\tan \left(10x^2\right)}{{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\frac{\tan \left(10\right)-10.\frac{\tan \left(10x^2\right)}{10x^2}}{\frac{{\sin }^{2}x}{x^2}}=\tan \left(10\right)-10$