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Question Number 193563 by SaRahAli last updated on 16/Jun/23

Answered by Gamil last updated on 16/Jun/23

Answered by Subhi last updated on 16/Jun/23

$$$$$${lim}_{x\to \frac{\pi }{4}}\frac{{sin}^3\left(x\right)-{cos}^3\left(x\right).(\frac{\pi }{4}-x)}{sin(\frac{\pi }{4}-x)(\frac{\pi }{4}-x)}$$$${lim}_{x\to \frac{\pi }{4}}\frac{{sin}^3\left(x\right)-{cos}^3\left(x\right)}{(\frac{\pi }{4}-x)}\Rrightarrow \frac{(sin\left(x\right)-cos\left(x\right))(({(sin\left(x\right)+cos\left(x\right))}^2-sin\left(x\right)cos\left(x\right))}{(\frac{\pi }{4}-x)}$$$${lim}_{x\to \frac{\pi }{4}}\frac{-cos\left(2x\right)(sin\left(x\right)+cos\left(x\right))}{(\frac{\pi }{4}-x)}+\frac{(cos\left(x\right)-sin\left(x\right)sin\left(2x\right)}{2(\frac{\pi }{4}-x)}.\frac{sin\left(x\right)+cos\left(x\right)}{sin\left(x\right)+cos\left(x\right)}$$$${lim}_{x\to \frac{\pi }{4}}\frac{-sin\left(2(\frac{\pi }{4}-x)\right)(sin\left(x\right)+cos\left(x\right))}{(\frac{\pi }{4}-x)}+\frac{sin(4(\frac{\pi }{4}-x)}{4(\frac{\pi }{4}-x).(sin\left(x\right)+cos\left(x\right))}$$$${lim}_{x\to \frac{\pi }{4}}-2(sin\left(x\right)+cos\left(x\right))+\frac{1}{sin\left(x\right)+cos\left(x\right)}=-2\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{-3\sqrt{2}}{2}$$

Answered by mnjuly1970 last updated on 16/Jun/23

$$$$$${lim}_{x\to \frac{\pi }{4}}\frac{{sin}^3\left(x\right)-{cos}^3\left(x\right)}{sin(\frac{\pi }{4}-x)}=?$$$${lim}_{x\to \frac{\pi }{4}}\frac{-\sqrt{2}sin(\frac{\pi }{4}-x)(1+\frac{1}{2}sin\left(2x\right))}{sin(\frac{\pi }{4}-x)}$$$$=-\sqrt{2}.(1+\frac{1}{2}sin\left(\frac{\pi }{2}\right))=-\frac{3\sqrt{2}}{2}$$$$hint:sin\left(x\right)-cos\left(x\right)=\sqrt{2}sin(x-\frac{\pi }{4})$$$${sin}^3\left(x\right)-{cos}^3\left(x\right)=(sin\left(x\right)-cos\left(x\right))(1+\frac{1}{2}sin\left(2x\right))$$$$◼$$$$$$

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