Question

If $L=\underset{x\to \pi }{lim}\frac{\tan \left(\pi {\cos }^{2}x\right)}{(x-\pi {)}^2}$,then $\frac{L}{\pi }$is equal to ?

• A. $-1$
• B. $-pi$
• C. $-2\pi$
• D. $-2$

Answer 1

The correct option is A $-1$

$L=\underset{x\to \pi }{lim}\frac{\tan \left(\pi {\cos }^{2}x\right)}{(x-\pi {)}^2}$

Let $x-\pi =y\Rightarrow x=y+\pi$

As $x\to \pi ,y\to 0$

$\therefore L=\underset{y\to 0}{lim}\frac{\tan \left(\pi {\cos }^2\right(y+x\left)\right)}{y^2}$

$=\underset{y\to 0}{lim}\frac{\tan \left(\pi {\cos }^{2}y\right)}{y^2}$

$=\underset{y\to 0}{lim}\frac{\tan (\pi -\pi {\sin }^{2}y)}{y^2}$

$=\underset{y\to 0}{lim}-\frac{\tan \left(\pi {\sin }^{2}y\right)}{y^2}$

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

$=-\pi \underset{y\to 0}{lim}\frac{\tan \left(\pi {\sin }^{2}y\right)}{\pi {\sin }^{2}y}\times \underset{y\to 0}{lim}\frac{{\sin }^{2}y}{y^2}$

$L=-\pi$

$\Rightarrow \frac{L}{\pi }=-1$.