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Question Number 192625 by Subhi last updated on 23/May/23

$${lim}_{x\to 0}\frac{{sin}^4\left(\pi cos\left(x\right)\right)}{1-cos(1-cos(1-cos\left(x\right)))}$$

Commented by Subhi last updated on 23/May/23

$$Igot8{\pi }^4$$

Answered by ARUNG_Brandon_MBU last updated on 23/May/23

$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{{\sin }^4\left(\pi \cos x\right)}{1-\cos (1-\cos (1-\cos x))}$$$$=\underset{x\to 0}{\lim }\frac{{\sin }^4(\pi -\frac{\pi x^2}{2})}{1-(1-\frac{x^8}{128})}[\sin t\underset{t\to 0}{\sim }t;\cos t\underset{t\to 0}{\sim }1-\frac{t^2}{2}]$$$$=\underset{x\to 0}{\lim }\frac{{\sin }^4\left(\frac{\pi x^2}{2}\right)}{\frac{x^8}{128}}=\underset{x\to 0}{\lim }\frac{\frac{{\pi }^4{x}^8}{16}}{\frac{x^8}{128}}=\overline{)\begin{array}{c}8{\pi }^4\end{array}}$$

Commented by Subhi last updated on 23/May/23

$$thanksboss$$$$Isolveditbythatway\downarrow \downarrow$$

Answered by Subhi last updated on 23/May/23

$$$$$${lim}_{x\to 0}\frac{{sin}^4\left(\pi cos\left(x\right)\right)}{1-cos(1-cos(1-cos\left(x\right)))}$$$$sin(\pi -\theta )=sin\left(\theta \right)$$$$x\to 0\Rrightarrow cos\left(x\right)\to 1\Rrightarrow \pi -\pi cos\left(x\right)\to 0$$$${lim}_{\pi -\pi cos\left(x\right)\to 0}\frac{{sin}^4(\pi -\pi cos\left(x\right)).{(\pi -\pi cos\left(x\right))}^4}{{(\pi -\pi cos\left(x\right))}^4.(1-cos(1-cos(1-cos\left(x\right)))}$$$$notethat{lim}_{x\to 0}\frac{sin\left(x\right)}{x}=1$$$${lim}_{x\to 0}\frac{{(\pi -\pi cos\left(x\right))}^4}{1-cos(1-cos(1-cos\left(x\right))}$$$$1-cos\left(x\right)=2{sin}^2\left(\frac{x}{2}\right)$$$$1-cos(1-cos(1-cos\left(x\right))=1-cos(1-cos\left(2{sin}^2\left(\frac{x}{2}\right)\right)$$$$=1-cos(2{sin}^2\left({sin}^2\left(\frac{x}{2}\right)\right)=2{sin}^2({sin}^2\left({sin}^2\left(\frac{x}{2}\right)\right)$$$${lim}_{x\to 0}\frac{{(\pi -\pi cos\left(x\right))}^4}{2{sin}^2\left({sin}^2\left({sin}^2\left(\frac{x}{2}\right)\right)\right)}$$$$x\to 0\Rrightarrow \frac{x}{2}\to 0\Rrightarrow {sin}^2\frac{x}{2}\to 0\Rrightarrow {sin}^2\left({sin}^2\left(\frac{x}{2}\right)\right)\to 0$$$${lim}_{{sin}^2\left({sin}^2\left(\frac{x}{2}\right)\right)\to 0}\frac{{(\pi -\pi cos\left(x\right))}^4.({sin}^2{\left({sin}^2\left(\frac{x}{2}\right)\right)}^2}{2{sin}^2\left({sin}^2\left({sin}^2\left(\frac{x}{2}\right)\right)\right).{sin}^4\left({sin}^2\left(\frac{x}{2}\right)\right)}$$$${lim}_{x\to 0}\frac{{(\pi -\pi cos\left(x\right))}^4}{2{sin}^4\left({sin}^2\left(\frac{x}{2}\right)\right)}$$$${lim}_{{sin}^2\left(\frac{x}{2}\right)\to 0}\frac{{(\pi -\pi cos\left(x\right))}^4.\left({sin}^8\left(\frac{x}{2}\right)\right)}{2{sin}^4\left({sin}^2\left(\frac{x}{2}\right)\right).{sin}^8\left(\frac{x}{2}\right)}$$$${lim}_{x\to 0}\frac{{\pi }^4{(1-cos\left(x\right))}^4}{2{sin}^8\left(\frac{x}{2}\right)}.\frac{8}{8}$$$$1-cos\left(x\right)=2{sin}^2\left(\frac{x}{2}\right)$$$${(1-cos\left(x\right))}^4=16{sin}^8\left(\frac{x}{2}\right)$$$$answer=8{\pi }^4$$$$$$

Answered by MM42 last updated on 24/May/23

$$Anothersolution$$$$\pi -\pi cosx=u\Rightarrow 1-cosx=\frac{u}{\pi }$$$$\Rightarrow L={lim}_{u\to 0}\frac{{sin}^4(\pi -u)}{1-cos(1-cos\left(\frac{u}{\pi }\right))}=$$$${lim}_{u\to 0}\frac{{sin}^4\left(u\right)}{1-cos\left(\frac{u^2}{2{\pi }^2}\right)}=$$$${lim}_{u\to 0}\frac{u^4}{\frac{u^4}{8{\pi }^2}}=8{\pi }^2✓$$

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