Question

The value of $\underset{x\to 1}{\lim }\frac{\left(x^2-1\right){\sin }^2(\pi x)}{x^4-2x^3+2x-1}$ is equal to:

• A. $\frac{{\pi }^2}{6}$
• B. $\frac{{\pi }^2}{3}$
• C. $\frac{{\pi }^2}{2}$
• D. ${\pi }^2$

Answer 1

Answer: (D)

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