Answer :

Solution :

$\mathrm{p}={\displaystyle \underset{\mathrm{x}\to \pi }{lim}\left(\frac{{\mathrm{x}}^2+\alpha \mathrm{x}+2{\pi }^2}{\mathrm{x}-\pi +2\sin \mathrm{x}}\right)}$

substitue x = π

, sin π = 0 We can see that denominator becomes zero.

Thus for p to have finite value numerator is also zero.

π² + απ + 2π² = 0

α = - 3π

$\mathrm{Now,\; p}=\left(\frac{{\pi }^2+\alpha \pi +2{\pi }^2}{\pi -\pi +2\sin \pi }\right)=\frac{0}{0}$ form

Using L'Hospital's rule, we get

$\mathrm{p}={\displaystyle \underset{\mathrm{x}\to \pi }{lim}\left(\frac{2\mathrm{x}+\alpha }{1+2\cos \mathrm{x}}\right)}$

substituting x = π, α = - 3π

$\mathrm{p}=\left(\frac{2\pi -3\pi }{1+2\cos \pi }\right)$

p = $\frac{-\pi }{-1}$

p = Π