Question

I need full explanation for the above question spamming will not be tolerated btw t thanks in advance :)​

Answer 1

Step-by-step explanation:

we have,

$\lim _{x \rarr \frac{\pi}{2} } \frac{(1 - \sin(x) )(8 {x}^{3} - {\pi}^{3}) \cos(x) }{(\pi - 2x)^{4} }$

$\lim _{x \rarr \frac{\pi}{2} } \frac{(1 - \sin(x))(2x - \pi)(4 {x}^{2} + {\pi}^{2} + 2\pi x) \cos(x) }{(2x - \pi)^{4} }$

$\lim _{x \rarr \frac{\pi}{2} } \frac{(1 - \sin(x))(4 {x}^{2} + 2\pi x + {\pi}^{2} ) \cos(x) }{(2x - \pi)^{3} }$

Let x - π/2 = y

=> 2x - π = 2y

$\lim _{y \rarr0 } \frac{(1 - \sin( \frac{\pi}{2} + y) )(4(y + \frac{\pi}{2} )^{2} + {\pi}^{2} + 2\pi(y + \frac{\pi}{2} )) \cos(y + \frac{\pi}{2} ) }{ {y}^{3} }$

$\lim _{y \rarr 0 } \frac{(1 - \cos(y)) (4(y + \frac{\pi}{2})^{2} + {\pi}^{2} + 2\pi(y + \frac{\pi}{2} ) (- \sin(y) )}{y^{3} }$

$- \lim _{y \rarr0 } \frac{2 \sin^{2} ( \frac{y}{2} )(4(y + \frac{\pi}{2} )^{2} + {\pi}^{2} + 2\pi(y + \frac{\pi}{2})) \sin(y) }{ {y}^{2} .y}$

$- \lim _{y \rarr0 } \frac{ \sin^{2} ( \frac{y}{2} ) (4(y + \frac{\pi}{2})^{2} + {\pi}^{2} + 2\pi(y + \frac{\pi}{2} )) \sin(y) }{y ^{2} .y}$

$- \lim _{y \rarr0 } \frac{ \sin^{2} ( \frac{y}{2} ) }{ {2( \frac{y}{2} )}^{2} } \times \lim _{y \rarr0 }(4(y + \frac{\pi}{2} )^{2} + {\pi}^{2} + 2\pi(y + \frac{\pi}{2} )) \times \lim _{y \rarr0 } \frac{ \sin(y) }{y}$

$= - \frac{1}{2} \times ( {\pi}^{2} + {\pi}^{2} + {\pi}^{2} ) \times 1$

$= - \frac{3 {\pi}^{2} }{2}$