Question

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

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

Answer 1

The correct option is C $4{\pi }^2$

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