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Question Number 107747 by ajfour last updated on 12/Aug/20

$$L=\underset{x\to 0}{\lim }\left(\frac{1-\cos x\cos 2x\cos 3x}{\sin {}^{2}2x}\right)=?$$

Answered by bobhans last updated on 12/Aug/20

$$\frac{\mathcal{B}\mathrm{obhans}}{\mathrm{\Pi }}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\left(\frac{1-(1-\frac{{\mathrm{x}}^2}{2})(1-{2\mathrm{x}}^2)(1-\frac{{9\mathrm{x}}^2}{2})}{{4\mathrm{x}}^2}\right)$$$$\mathrm{L}=\underset{x\to 0}{\lim }\left(\frac{1-(1-\frac{5}{2}{\mathrm{x}}^2+{\mathrm{x}}^4)(1-\frac{9}{2}{\mathrm{x}}^2)}{{4\mathrm{x}}^2}\right)$$$$\mathrm{L}=\underset{x\to 0}{\lim }\left(\frac{1-(1-{7\mathrm{x}}^2+\frac{45}{4}{\mathrm{x}}^4-\frac{9}{2}{\mathrm{x}}^4)}{{4\mathrm{x}}^2}\right)=\frac{7}{4}$$

Commented by ajfour last updated on 12/Aug/20

$$ThanksSir,mostconvinientway$$$$thisis..$$

Commented by malwaan last updated on 12/Aug/20

$$SirBobhans$$$$youcandelet{x}^{4}termsbecause$$$$weneedonly{x}^{2}so$$$$\mathrm{L}=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{1-(1-{7\mathrm{x}}^2+...)}{{4\mathrm{x}}^2}\right)=\frac{7}{4}$$$$thankyouSir$$

Answered by john santu last updated on 12/Aug/20

$$\frac{\divideontimes \mathcal{JS}\divideontimes }{\mathrm{\Sigma }}$$$$ingenerrally$$$$\underset{x\to 0}{\lim }\frac{1-\cos px\cos qx\cos rx\cos tx}{\sin {}^{2}ax}$$$$=\frac{\frac{1}{2}(p^2+q^2+r^2+t^2)}{a^2}$$$$now\underset{x\to 0}{\lim }\frac{1-\cos x\cos 2x\cos 3x}{\sin {}^{2}2x}=$$$$\frac{\frac{1}{2}(1+4+9)}{4}=\frac{\frac{1}{2}.14}{4}=\frac{7}{4}$$

Commented by bemath last updated on 12/Aug/20

$$joss$$

Commented by ajfour last updated on 12/Aug/20

$$Thanksforthegeneralformula,$$$$canipleasehaveaproofofit..$$

Commented by malwaan last updated on 12/Aug/20

$$SirJohnsantu$$$$isthisright$$$$\underset{x\to 0}{lim}\frac{1-cos{a}_{1}xcos{a}_{2}x....cos{a}_{n}x}{{sin}^{2}bx}$$$$=\frac{\frac{1}{2}(a_1^2+a_2^2+...+a_n^2)}{b^2}$$$$n⩾1$$$$thankyousomuch$$

Answered by mathmax by abdo last updated on 12/Aug/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{1-\mathrm{cosx}\cos \left(2\mathrm{x}\right)\cos \left(3\mathrm{x}\right)}{{\sin }^2\left(2\mathrm{x}\right)}\mathrm{we}\mathrm{hsve}$$$$\mathrm{cosx}\cos \left(2\mathrm{x}\right)\cos \left(3\mathrm{x}\right)=\frac{1}{2}(\cos \left(3\mathrm{x}\right)+\cos \left(\mathrm{x}\right)\cos \left(3\mathrm{x}\right)$$$$=\frac{1}{2}{\cos }^2\left(3\mathrm{x}\right)+\frac{1}{4}(\cos \left(4\mathrm{x}\right)+\cos \left(2\mathrm{x}\right))$$$$=\frac{1}{4}(1+\cos \left(6\mathrm{x}\right)+\cos \left(4\mathrm{x}\right)+\cos \left(2\mathrm{x}\right))\sim$$$$\frac{1}{4}(1+1-\frac{{\left(6\mathrm{x}\right)}^2}{2}+1-\frac{{\left(4\mathrm{x}\right)}^2}{2}+1-\frac{{\left(2\mathrm{x}\right)}^2}{2})$$$$=\frac{1}{4}(4-{18\mathrm{x}}^2-{8\mathrm{x}}^2-{2\mathrm{x}}^2)=\frac{1}{4}(4-{28\mathrm{x}}^2)=1-{7\mathrm{x}}^2$$$${\sin }^2\left(2\mathrm{x}\right)=\frac{1-\cos \left(4\mathrm{x}\right)}{2}=\frac{1}{2}-\frac{1}{2}\cos \left(4\mathrm{x}\right)\sim \frac{1}{2}-\frac{1}{2}(1-\frac{{\left(4\mathrm{x}\right)}^2}{2})$$$$={4\mathrm{x}}^2\mathrm{and}1-\mathrm{cosx}\cos \left(2\mathrm{x}\right)\cos \left(3\mathrm{x}\right)\sim {7\mathrm{x}}^2\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{7\mathrm{x}}^2}{{4\mathrm{x}}^2}=\frac{7}{4}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{7}{4}$$$$$$

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