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Question Number 119725 by bemath last updated on 26/Oct/20

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{4\sqrt{2}-{(\cos x+\sin x)}^5}{{(\cos x-\sin x)}^2}=?$$

Commented by bemath last updated on 26/Oct/20

$$comparevia{L}^{\prime }Hopital$$$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{-5{(\cos x+\sin x)}^4(-\sin x+\cos x)}{2(\cos x-\sin x)(-\cos x-\sin x)}$$$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{-5{(\cos x+\sin x)}^4(\cos x-\sin x)}{-2(\cos x-\sin x)(\cos x+\sin x)}=$$$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{5}{2}{(\cos x+\sin x)}^3=$$$$\frac{5}{2}{\left(\sqrt{2}\right)}^3=\frac{5}{2}\times 2\sqrt{2}=5\sqrt{2}$$

Answered by benjo_mathlover last updated on 26/Oct/20

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{4\sqrt{2}-4\sqrt{2}\sin {}^5(x+\frac{\pi }{4})}{1-\sin 2x}=$$$$[letx=\frac{\pi }{4}+z]$$$$\underset{z\to 0}{\lim }\frac{4\sqrt{2}(1-\sin {}^5(z+\frac{\pi }{2}))}{1-\sin (2z+\frac{\pi }{2})}=$$$$\underset{z\to 0}{\lim }\frac{4\sqrt{2}(1-\cos {}^{5}z)}{1-\cos 2z}=\underset{z\to 0}{\lim }\frac{4\sqrt{2}(1-{(1-\frac{z^2}{2})}^5)}{1-(1-\frac{4z^2}{2})}$$$$=\underset{z\to 0}{\lim }\frac{4\sqrt{2}(1-(1-\frac{5z^2}{2}))}{2z^2}=\underset{z\to 0}{\lim }\frac{2\sqrt{2}\left(\frac{5z^2}{2}\right)}{z^2}=5\sqrt{2}$$$$$$$$$$

Answered by Dwaipayan Shikari last updated on 26/Oct/20

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{4\sqrt{2}(1-{cos}^5(\frac{\pi }{4}-x))}{{(cosx-sinx)}^2}$$$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{4\sqrt{2}(1-{cos}^5(\frac{\pi }{4}-x))}{2{cos}^2(\frac{\pi }{4}+x)}$$$$\underset{x\to \frac{\pi }{4}}{\lim 2}\sqrt{2}\frac{1-{cos}^5(\frac{\pi }{4}-x)}{{sin}^2(\frac{\pi }{4}-x)}$$$$\underset{x\to \frac{\pi }{4}}{\lim 2}\sqrt{2}(1+cos(\frac{\pi }{4}-x)+{cos}^2(\frac{\pi }{4}-x)+{cos}^3(\frac{\pi }{4}-x)+{cos}^4(\frac{\pi }{4}-x))\frac{1-cos(\frac{\pi }{4}-x)}{{sin}^2(\frac{\pi }{4}-x)}$$$$2\sqrt{2}(1+1+1+1+1)\frac{2{(\frac{\pi }{4}-x)}^2}{4.{(\frac{\pi }{4}-x)}^2}=5\sqrt{2}$$$$$$

Answered by Bird last updated on 26/Oct/20

$$f\left(x\right)=\frac{4\sqrt{2}-{(cosx+sinx)}^5}{{(cosx-sinx)}^2}$$$$\Rightarrow f\left(x\right)=\frac{4\sqrt{2}-(\sqrt{2}cos{(x-\frac{\pi }{4})}^5}{{\left(\sqrt{2}cos(x+\frac{\pi }{4})\right)}^2}$$$$=\frac{4\sqrt{2}-4\sqrt{2}{cos}^5(x-\frac{\pi }{4})}{2{cos}^2(x+\frac{\pi }{4})}$$$${=}_{x-\frac{\pi }{4}=t}2\sqrt{2}\times \frac{1-{cos}^{5}t}{{cos}^2(t+\frac{\pi }{2})}=f(t+\frac{\pi }{4})$$$$$$$$=2\sqrt{2}\times \frac{1-{cos}^{5}t}{{sin}^{2}t}$$$$=2\sqrt{2}\times \frac{(1-cost)(1+cost+{cos}^{2}t+{cos}^{3}t+{cos}^{4}t)}{{sin}^{2}t}$$$$f(t+\frac{\pi }{4})\sim 2\sqrt{2}\times \frac{\frac{t^2}{2}}{t^2}(1+cost+...+{cos}^{4}t)$$$$=\sqrt{2}(1+cost+...+{cos}^{4}t)\Rightarrow$$$${lim}_{t\to 0}f(t+\frac{\pi }{4})=5\sqrt{2}={lim}_{x\to \frac{\pi }{4}}f\left(x\right)$$$$$$

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