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. $1$

Answer 1

Answer: (B)

$\pi$

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}$ $\left(\frac{0}{0}form\right)$
$=\underset{x\to 0}{lim}\frac{\left\{\cos \right(\pi {\cos }^{2}x)\cdot \pi \cdot 2\cos x(-\sin x\left)\right\}}{2x}$
(using L'Hospital's rule)
$=\underset{x\to 0}{lim}\pi \cos \left(\pi {\cos }^{2}x\right)\cdot \cos x\cdot \left(\frac{-\sin x}{x}\right)$
$=\left(\pi \right)\cdot (-1)\cdot 1\cdot (-1)=\pi$.