find-lim-x-pi-2-sinx-ln-x-pi-2-

Question Number 109191 by abdomsup last updated on 21/Aug/20

f i n d {l i m}_{x \to \frac{π}{2}} {(s i n x)}^{l n ∣ x - \frac{π}{2} ∣}

Commented by bemath last updated on 22/Aug/20

s e t x = \frac{π}{2} + z

L = \lim_{z \to 0} {(\sin (\frac{π}{2} + z))}^{\ln ∣ z ∣} = \lim_{z \to 0} {(\cos z)}^{\ln ∣ z ∣}

\ln L = \lim_{z \to 0} \ln ∣ z ∣ (\cos z)

\ln L = \lim_{z \to 0} \frac{\cos z}{\frac{1}{\ln z}} = \lim_{z \to 0} \frac{- \sin z}{[\frac{- \frac{1}{z}}{\ln^{2} (z)}]}

\ln L = \lim_{z \to 0} \sin z . \frac{\ln^{2} z}{\frac{1}{z}} = \lim_{z \to 0} z . \sin z . \ln^{2} z

\ln L = 0 \Rightarrow L = e^{0} = 1

Answered by mathmax by abdo last updated on 23/Aug/20

f (x) = {(sinx)}^{\ln ∣ x - \frac{π}{2} ∣} \Rightarrow f (x) = e^{\ln ∣ x - \frac{π}{2} ∣ \ln (sinx)} changement x - \frac{π}{2} = t

give f (x) = g (t) = e^{\ln ∣ t ∣ \ln (\sin (\frac{π}{2} + t))} = e^{\ln ∣ t ∣ \ln (cost)}

(x \to \frac{π}{2} \Rightarrow t \to 0) we have \ln (cost) = \ln (\cos ∣ t ∣) \sim \ln (1 - \frac{∣ t ∣^{2}}{2}) \Rightarrow

\sim - \frac{∣ t ∣^{2}}{2} \Rightarrow \ln ∣ t ∣ \ln (cost) \sim - \ln ∣ t ∣ \frac{∣ t ∣^{2}}{2} \to 0 (t \to 0) \Rightarrow

\lim_{t \to 0} g (t) = 1 = \lim_{x \to \frac{π}{2}} f (x)