Question

$\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\pi {\cos }^{4}x\right)}{x^4}$ is equal to :

• A. ${\pi }^2$
• B. $2{\pi }^2$
• C. $4{\pi }^2$
• D. $4\pi$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\pi {\cos }^{4}x\right)}{x^4}$
$\underset{x\to 0}{\lim }\frac{1-\cos \left(2\pi {\cos }^{4}x\right)}{2x^4}$
$\underset{x\to 0}{\lim }\frac{1-\cos \left(2\pi -2\pi {\cos }^{4}x\right)}{{\left[2\pi \left(1-{\cos }^{4}x\right)\right]}^2}4{\pi }^2\cdot \frac{{\sin }^{4}x}{2x^4}{\left(1+{\cos }^{2}x\right)}^2$
$=\frac{1}{2}\cdot 4{\pi }^2\cdot \frac{1}{2}(2{)}^2=4{\pi }^2$