Question

$\underset{x\to 0}{\lim }\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}$ is equal to

• A. - $\pi$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. 1
• E. $\frac{1}{\pi }$

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}$
$=\underset{x\to 0}{\lim }\left\{\frac{\cos \left(\pi {\cos }^{2}x\right)\cdot \pi \cdot 2\cos x(-\sin x)}{2x}\right\}$ (using $L^{{}^{\prime }}$ Hopitals, Rule)
$=\underset{x\to 0}{\lim }\pi \cos \left(\pi {\cos }^{2}x\right)\cdot \cos x\cdot \left(\frac{-\sin x}{x}\right)$
$=\pi (-1).1.(-1)=\pi$