Question

$\displaystyle \lim_{x\to0} \frac{sin\left(\pi\,cos^{2} x\right)}{x^{2}}$ equals

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

Answer 1

Answer: (D)

Consider
$\displaystyle \lim_{x\to0} \frac{sin\left(\pi\,cos^{2} x\right)}{x^{2}}$
$= \displaystyle \lim_{x\to0} \frac{sin\left(\pi-\pi\,sin^{2} x\right)}{x^{2}} $
$\left[\because sin \left(\pi - \theta\right) = sin \,\theta\right]$
$= \displaystyle \lim _{x\to 0} \frac{sin\left(\pi \,sin^{2} x\right)}{\pi \,sin^{2}x}\times\frac{\left(\pi \,sin^{2} x\right)}{x^{2}} = \pi$