Question

$ \displaystyle\lim_{ x \to 0 } \frac{ sin ( \pi \, cos^2 \, x)}{ x^2} $ is equal to

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

Answer 1

Answer: (D)

$\displaystyle\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}} $
$=\displaystyle\lim _{x \rightarrow 0} \frac{\sin \left(\pi\left(1-\sin ^{2} x\right)\right.}{x^{2}} $
$=\displaystyle\lim _{x \rightarrow 0} \frac{ \sin\left(\pi-\pi \sin ^{2} x\right)}{x^{2}} $
$=\displaystyle\lim _{x \rightarrow 0} \frac{ \sin\left(\pi \sin ^{2} x\right)}{x^{2}} \quad[\because \sin (\pi-\theta)=\sin \theta] $
$=\displaystyle\lim _{x \rightarrow 0} \frac{ \sin\left(\pi \sin ^{2} x\right)}{\left(\pi \sin ^{2} x\right)} \times \frac{\pi \sin ^{2} x}{x^{2}} $
$=\displaystyle\lim _{x \rightarrow 0} 1 \times \pi\left(\frac{\sin x}{x}\right)^{2}=\pi$