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Question Number 73117 by aliesam last updated on 06/Nov/19

Commented by mathmax by abdo last updated on 06/Nov/19

$$wehavecos\left(3x\right)\sim 1-\frac{{\left(3x\right)}^2}{2}(x\to 0)\Rightarrow cos\left(3x\right)-1\sim -\frac{9x^2}{2}\Rightarrow$$$$1-cos\left(3x\right)\sim \frac{9x^2}{2}\Rightarrow \frac{1-cos\left(3x\right)}{x^2}\sim \frac{9}{2}\Rightarrow {lim}_{x\to 0}\frac{1-cos\left(3x\right)}{x^2}=\frac{9}{2}$$

Answered by mind is power last updated on 06/Nov/19

$$\cos \left(3\mathrm{x}\right)={4\cos }^3\left(\mathrm{x}\right)-3\cos \left(\mathrm{x}\right)$$$$1-\cos \left(3\mathrm{x}\right)=-{4\cos }^3\left(\mathrm{x}\right)+3\cos \left(\mathrm{x}\right)+1=-4(\cos \left(\mathrm{x}\right)-1)({\cos }^2\left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)+\frac{1}{4})$$$$=-4(-{2\sin }^2\left(\frac{\mathrm{x}}{2}\right))({\cos }^2\left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)+\frac{1}{4})$$$$\frac{1-\cos \left(3\mathrm{x}\right)}{{\mathrm{x}}^2}=\frac{{8\sin }^2\left(\frac{\mathrm{x}}{2}\right)}{4{\left(\frac{\mathrm{x}}{2}\right)}^2}.({\cos }^2\left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)+\frac{1}{4})$$$$=2.{\left(\frac{\sin \left(\frac{\mathrm{x}}{2}\right)}{\frac{\mathrm{x}}{2}}\right)}^2.({\cos }^2\left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)+\frac{1}{4})$$$$\underset{x\to 0}{\lim }\frac{\sin \left(\mathrm{x}\right)}{\mathrm{x}}=1$$$$\Rightarrow =2.1^2.(1+1+\frac{1}{4})=\frac{9}{2}$$

Answered by malwaan last updated on 07/Nov/19

$$\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{[1-\mathit{c}\mathit{o}\mathit{s}\left(3\mathit{x}\right)]\times [1+\mathit{c}\mathit{o}\mathit{s}\left(3\mathit{x}\right)]}{{\mathit{x}}^2\times [1+\mathit{c}\mathit{o}\mathit{s}\left(3\mathit{x}\right)]}$$$$=\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{{\mathit{s}\mathit{i}\mathit{n}}^2\left(3\mathit{x}\right)}{{\mathit{x}}^2[1+\mathit{c}\mathit{o}\mathit{s}\left(3\mathit{x}\right)]}=\frac{3^2}{1+1}=\frac{9}{2}$$

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