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

The correct option is A 0

${lim}_{x\to 1}\frac{1+\sin \pi \left(\frac{3x}{1+x^2}\right)}{1+\cos \pi x}\frac{0}{0}$form

$={lim}_{x\to 1}\frac{\cos \pi \left(\frac{3x}{1+x^2}\right)}{-\sin \pi x\times \pi }\times \left[\frac{(1+x^2)3-3x\left(2x\right)}{{(1+x^2)}^2}\right]\times \frac{\pi x}{\pi x}$

$={lim}_{x\to 1}-\frac{\cos \pi \left(\frac{3x}{1+x^2}\right)\left[\frac{3(1+x^2)-6x^2}{{(1+x^2)}^2}\right]}{{\pi }^{2}x}[as\pi x\to \pi ;{lim}_{\pi x\to \pi }\frac{\sin \pi x}{\pi x}=1]$

$=-\frac{0}{{\pi }^2}\left[\frac{3\left(2\right)-6}{2^2}\right]=0$