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Question Number 144450 by mnjuly1970 last updated on 25/Jun/21

$$$$$$.........\mathrm{C}alculus\left(\mathrm{I}\right).........$$$${\mathrm{Lim}}_{x\to 0}\frac{1-cos(xcos\left(\frac{x}{2}\right).cos\left(\frac{x}{4}\right)cos\left(\frac{x}{8}\right))}{x^2}=?$$

Answered by Dwaipayan Shikari last updated on 25/Jun/21

$$\underset{x\to 0}{\lim }\frac{1-cos\left(xcos\left(x/2\right)cos\left(x/4\right)cos\left(x/8\right)\right)}{x^2}=y$$$$log(1-x^{2}y)=log\left(cos(xcos\left(\frac{x}{2}\right)..cos\left(\frac{x}{8}\right))\right)\underset{x\to 0}{\lim }log(1+x)\approx x$$$$\underset{x\to 0}{\lim }-x^{2}y\approx -\frac{x^2{cos}^2\left(\frac{x}{2}\right){cos}^2\left(\frac{x}{4}\right){cos}^2\left(\frac{x}{8}\right)}{2!}$$$$y=\frac{1}{2}$$

Commented by mnjuly1970 last updated on 25/Jun/21

$$grateful...$$

Answered by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{L}={\lim }_{\mathrm{x}\to 0}\frac{1-\mathrm{cosx}.\cos \left(\frac{\mathrm{x}}{4}\right).\cos \left(\frac{\mathrm{x}}{8}\right)}{{\mathrm{x}}^2}\mathrm{we}\mathrm{have}$$$$\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\mathrm{and}\cos \left(\frac{\mathrm{x}}{4}\right)\sim 1-\frac{{\mathrm{x}}^2}{32}$$$$\cos \left(\frac{\mathrm{x}}{8}\right)\sim 1-\frac{{\mathrm{x}}^2}{128}\Rightarrow \mathrm{L}\sim \frac{1-(1-\frac{{\mathrm{x}}^2}{2})(1-\frac{{\mathrm{x}}^2}{32})(1-\frac{{\mathrm{x}}^2}{128})}{{\mathrm{x}}^2}$$$$=\frac{1-(1-\frac{{\mathrm{x}}^2}{32}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{64})(1-\frac{{\mathrm{x}}^2}{128})}{{\mathrm{x}}^2}$$$$=\frac{1-(1-\frac{17}{32}{\mathrm{x}}^2+\frac{{\mathrm{x}}^4}{64})(1-\frac{{\mathrm{x}}^2}{128})}{{\mathrm{x}}^2}$$$$=\frac{1-(1-\frac{{\mathrm{x}}^2}{128}-\frac{17}{32}{\mathrm{x}}^2+\frac{17}{32.128}{\mathrm{x}}^4+\frac{{\mathrm{x}}^4}{64}-\frac{{\mathrm{x}}^6}{64.128})}{{\mathrm{x}}^2}$$$$=\frac{1}{128}+\frac{17}{32}+(...){\mathrm{x}}^2+(...){\mathrm{x}}^4\Rightarrow$$$${\mathrm{Lim}}_{\mathrm{x}\to 0}=\frac{1}{128}+\frac{68}{128}=\frac{69}{128}$$

Answered by mnjuly1970 last updated on 26/Jun/21

$$solution:$$$${\mathrm{Lim}}_{x\to 0}\frac{1-cos(x\left\{cos\left(\frac{x}{2}\right)cos\left(\frac{x}{4}\right)cos\left(\frac{x}{8}\right)\right\}=f\left(x\right))}{x^2}$$$$:=\frac{{\mathrm{Lim}}_{x\to 0}\left(1\right)-cos\left({\mathrm{Lim}}_{x\to 0}xf\left(x\right)\right)}{{\mathrm{Lim}}_{x\to 0}\left(x^2\right)}$$$$:\underset{asx\to 0}{\stackrel{f\left(x\right)\to 0}{=}}{\mathrm{Lim}}_{x\to 0}\frac{1-cos\left(x\right)}{x^2}={\mathrm{Lim}}_{x\to 0}\frac{2{sin}^2\left(\frac{x}{2}\right)}{x^2}$$$$:=\frac{1}{2}{\mathrm{Lim}}_{x\to 0}{\left(\frac{sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)}^2=1$$$$....m.n.july.1970....$$

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