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Question Number 66316 by mathmax by abdo last updated on 12/Aug/19

$${lim}_{x\to \frac{\pi }{2}}\frac{ln\left({sin}^{2}x\right)}{{(\frac{\pi }{2}-x)}^2}$$$$$$

Commented by mathmax by abdo last updated on 21/Aug/19

$$letf\left(x\right)=\frac{ln\left({sin}^{2}x\right)}{{(\frac{\pi }{2}-x)}^2}changement\frac{\pi }{2}-x=tgive$$$${lim}_{x\to \frac{\pi }{2}}f\left(x\right)={lim}_{t\to 0}\frac{ln\left({cos}^{2}t\right)}{t^2}={lim}_{t\to 0}\frac{ln\left(\frac{1+cos\left(2t\right)}{2}\right)}{t^2}$$$$butln\left(\frac{1+cos\left(2t\right)}{2}\right)=ln(1+cos\left(2t\right))-ln\left(2\right)$$$$cos\left(2t\right)\sim 1-2t^2+o\left(t^2\right)\Rightarrow 1+cos\left(2t\right)\sim 2-2t^2\Rightarrow$$$$ln(1+cos\left(2t\right))\sim ln\left(2\right)+ln(1-t^2)\sim ln\left(2\right)-t^2\Rightarrow$$$$\frac{ln\left(\frac{1+cos\left(2t\right)}{2}\right)}{t^2}\sim -1(t\to o)\Rightarrow {lim}_{x\to \frac{\pi }{2}}f\left(x\right)=-1$$

Answered by kaivan.ahmadi last updated on 12/Aug/19

$$\stackrel{hop}{=}{lim}_{x\to \frac{\pi }{2}}\frac{sin2x}{-2{sin}^{2}x(\frac{\pi }{2}-x)}\stackrel{hop}{=}$$$${lim}_{x\to \frac{\pi }{2}}\frac{2cos2x}{-2sin2x(\frac{\pi }{2}-x)+2{sin}^{2}x}=$$$$\frac{-2}{2}=-1$$

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