Question

$\underset{x\to 1}{\lim }\frac{1+\sin \pi \left(\frac{3x}{1+x^2}\right)}{1+\cos \pi x}$ is equal to

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A)

$\underset{x\to 1}{\lim }\frac{1+\sin \pi \left(\frac{3x}{1+x^2}\right)}{1+\cos \pi x}$
$=\underset{x\to 1}{\lim }\frac{1-\cos \left(\frac{3\pi }{2}-\frac{3\pi x}{1+x^2}\right)}{1-\cos (\pi -\pi x)}$
$=\underset{x\to 1}{\lim }\frac{2{\sin }^2\left(\frac{3\pi }{4}-\frac{3\pi x}{2\left(1+x^2\right)}\right)}{2{\sin }^2\left(\frac{\pi }{2}-\frac{\pi x}{2}\right)}$
$=\underset{x\to 1}{\lim }{\left(\frac{\frac{3\pi }{4}-\frac{3\pi x}{2\left(1+x^2\right)}}{\frac{\pi }{2}-\frac{\pi x}{2}}\right)}^2$
$=\underset{x\to 1}{\lim }9{\left(\frac{\frac{1}{2}-\frac{x}{1+x^2}}{1-x}\right)}^2$
$=\underset{x\to 1}{\lim }9{\left(\frac{x-1}{2\left(1+x^2\right)}\right)}^2=0$