Question

The value of lim x—pi/2 [xtanx-(pi/2)secx]

Answer 1

❏Question :-

$\sf \: find \: the \: value \: of \: \\ \: \lim_{x \to \frac{ \pi}{2} } \{x \tan x - \big( \frac{ \pi}{2} \big) \sec x \} \:$

❏Answer :-

$\boxed{\lim_{x \to \frac{ \pi}{2} } \{x \tan x - \big( \frac{ \pi}{2} \big) \sec x \} \: = \frac{ { \pi}^{2} }{4} - \frac{1}{2} } \:$

❏Used property :-

$\boxed{ \sf\lim \frac{ \sin h}{h} = 1} \\ \boxed{ \sf\lim \frac{ \tan h}{h} = 1} \:$

❏Solution :-

$\leadsto \: \lim_{x \to \frac{ \pi}{2} } \{x \tan x - \big( \frac{ \pi}{2} \big) \sec x \} \: \: \\ \\ \leadsto \: \lim_{x \to \frac{ \pi}{2} } \{x \tan x \big( \frac{x}{x} \big)- \big( \frac{ \pi}{2} \big) \sec x \} \: \\ \\ \because \: \sec x = \frac{1}{ \cos x} \: \\ \therefore\\ \leadsto \: \lim_{x \to \frac{ \pi}{2} } \{ {x}^{2} \big( \frac{ \tan x}{x} \big)- \big( \frac{ \pi}{2} \big) \frac{1}{ \cos x} \} \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \because \: \: \boxed{ \sf\lim \frac{ \tan h}{h} = 1} \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\sin( \frac{ \pi}{2} + \theta) = \cos \theta}\\ \therefore \\ \leadsto \: \lim_{x \to \frac{ \pi}{2} } \{ {x}^{2} - \big( \frac{ \pi}{2} \big) \frac{1}{ \sin( \frac{ \pi}{2} + x)} \} \: \\ \\ \leadsto \: \lim_{x \to \frac{ \pi}{2} } \{ {x}^{2} - \big( \frac{ \pi}{2} \big) \frac{1}{ \sin( \frac{ \pi}{2} + x)} \bigg( \frac{ \frac{ \pi}{2} + x }{ \frac{ \pi}{2} + x } \bigg) \} \: \: \\ \\ \leadsto \: \lim_{x \to \frac{ \pi}{2} } \{ {x}^{2} - \big( \frac{ \pi}{2} \big) \frac{1}{ ( \frac{ \pi}{2} + x)} \} \: \: \\ \\ \leadsto \: \lim_{x \to \frac{ \pi}{2} } \{ {x}^{2} - \frac{ \pi}{ ( \pi + 2x)} \} \: \\ \\ \mathbb{Taking \: Limit \: } \\ \\ \leadsto \: { \bigg( \frac{ \pi}{2} \bigg)}^{2} - \frac{ \pi}{ \pi + \pi} \\ \\ \leadsto \: \frac{ { \pi}^{2} }{4} - \frac{1}{2} \\ \\ \boxed{\lim_{x \to \frac{ \pi}{2} } \{x \tan x - \big( \frac{ \pi}{2} \big) \sec x \} \: = \frac{ { \pi}^{2} }{4} - \frac{1}{2}}$