Question

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

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

Answer 1

Answer: (B)

$\displaystyle\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}$
$=\displaystyle\lim _{x \rightarrow 0}\left\{\frac{\cos \left(\pi \cos ^{2} x\right) \cdot \pi \cdot 2 \cos x(-\sin x)}{2 x}\right\}$ (using $L^{'}$ Hopitals, Rule)
$=\displaystyle\lim _{x \rightarrow 0} \pi \cos \left(\pi \cos ^{2} x\right) \cdot \cos x \cdot\left(\frac{-\sin x}{x}\right)$
$=\pi(-1) . 1 .(-1)=\pi$