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Question Number 34540 by rahul 19 last updated on 07/May/18

$$a)\underset{x\to \frac{\pi }{4}}{\lim }\frac{{(\cos x+\sin x)}^3-2\sqrt{2}}{1-\sin 2x}=?$$$$b)\underset{x\to 0}{\lim }{\left(\sin x\right)}^{\frac{1}{x}}=?$$

Commented by abdo mathsup 649 cc last updated on 07/May/18

$$letputf\left(x\right)=\frac{{(cosx+sinx)}^3-2\sqrt{2}}{1-sin\left(2x\right)}$$$$f\left(x\right)=\frac{{\left\{\sqrt{2}cos(x-\frac{\pi }{4})\right\}}^3-2\sqrt{2}}{1-sin\left(2x\right)}.changement$$$$x-\frac{\pi }{4}=tgivex=\frac{\pi }{4}+tand$$$$lim{}_{x\to \frac{\pi }{4}}f\left(x\right)={lim}_{t\to 0}\frac{2\sqrt{2}{cos}^{3}t-2\sqrt{2}}{1-sin(\frac{\pi }{2}+2t)}$$$$={lim}_{t\to 0}2\sqrt{2}\frac{{cos}^{3}t-1}{1-cos\left(2t\right)}$$$$={lim}_{t\to 0}-2\sqrt{2}\frac{(1-cost)(1+cost+{cos}^{2}t)}{1-cos\left(2t\right)}but$$$$1-cost\sim \frac{t^2}{2}and1-cos\left(2t\right)\sim 2t^2(t\in v\left(0\right))\Rightarrow$$$$\frac{(1-cost)}{1-cos\left(2t\right)}\sim \frac{\frac{t^2}{2}}{2t^2}=\frac{1}{4}\Rightarrow {lim}_{x\to \frac{\pi }{4}}f\left(x\right)=-2\sqrt{2}\frac{1}{4}.3$$$${lim}_{x\to \frac{\pi }{4}}f\left(x\right)=-3\frac{\sqrt{2}}{2}.$$

Commented by abdo mathsup 649 cc last updated on 08/May/18

$${\left(sinx\right)}^{\frac{1}{x}}=e^{\frac{ln\left(sinx\right)}{x}}but\frac{ln\left(sinx\right)}{x}\sim \frac{ln\left(x\right)}{x}{}_{x\to 0^{+}}\to -\mathrm{\infty }$$$$\Rightarrow {lim}_{x\to 0^{+}}{\left(sinx\right)}^{\frac{1}{x}}=0.$$

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