lim-x-pi-2-1-sin-pi-2x-5-sin-x-pi-2-ln-4-cos-pi-2x-1-pi-2-x-2-

Question Number 170018 by cortano1 last updated on 14/May/22

\lim_{x \to \frac{π}{2}} {(1 + \sin (π - 2 x))}^{\frac{5}{\sin (x - \frac{π}{2})}} . \ln (4 . \frac{\cos (π - 2 x) - 1}{{(\frac{π}{2} - x)}^{2}}) = ?

Answered by Mathspace last updated on 14/May/22

u (x) = {(1 + s i n (π - 2 x))}^{\frac{5}{s i n (x - \frac{π}{2})}}

a n d v (x) = l n (4 . \frac{c o s (π - 2 x) - 1}{{(\frac{π}{2} - x)}^{2}}

\Rightarrow u (x) = {(1 + s i n (2 x))}^{\frac{5}{- c o s x}}

= e^{- \frac{5}{c o s x} l n (1 + s i n (2 x)}

c h . x - \frac{π}{2} = t g i v e

u (x) = u (\frac{π}{2} + t) = e^{- \frac{5}{- s i n t} l n (1 - s i n (2 t))}

s i n t \sim t a n d s i n (2 t) \sim 2 t \Rightarrow

l n (1 - s i n (2 t)) \sim l n (1 - 2 t) \sim - 2 t \Rightarrow

\frac{5}{s i n t} l n (1 - s i n (2 t) \sim \frac{5}{t} (- 2 t) = - 10 \Rightarrow

l i m u (x) = e^{- 10}

v (x) = l n (4 . \frac{- c o s (2 x) - 1}{{(\frac{π}{2} - x)}^{2}}) (x - \frac{π}{2} = t)

= l n (4 \times \frac{- c o s (2 (\frac{π}{2} + t) - 1}{t^{2}})

= l n (- 4 . \frac{- c o s (2 t) + 1}{t^{2}})

= l n (\frac{4}{t^{2}} (c o s (2 t) - 1))

b u t c o s (2 t) - 1 < 0 e r r o r a t t h e q u e s t i o n! \dots

c o s (2 t) \sim 1 - 2 t^{2} \Rightarrow 1 - c o s (2 t)