InterviewSolution
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InterviewSolution
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`lim_(x to 0) ((1-cos x cos2x cos3x)/(sin ^(2)2x))` का मान ज्ञात कीजिए। |
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Answer» हम जानते है कि `cos x cos2x cos 3x=1/2(2 cos x cos3x cos2x)` `=1/2[(cos 2x+cos 4x)cos2x]` `=1/4[(2 cos ^(2)2x+2 cos 4x cos 2x)]` `=1/4[1+cos 4x+cos 2x+cos 6x]` इसलिए `underset(x to0)lim[(1-cos x cos2x cos 3x)/(sin^(2)2x)]=underset(xto0)lim[(1-(1)/(4)(1+cos 4x+cos 2x+cos 6x))/(sin^(2)2x)]` `=underset(xto0)lim[(1-cos 2x+1-cos4x+1-cos 6x)/(4 sin^(2)2x)]` `=underset(x to 0)lim(2 sin ^(2)x+2sin ^(2)2x+2sin^(2)3x)/(4sin^(2)2x)` `=underset(x to 0)lim{(2((sin x)/(x))^(2)*x^(2)+2((sin 2x)/(2x))^(2)*4x^(2)+2((sin 3x)/(3x))^(2)*9x^(2))/(4((sin 2x)/(2x))*4x^(2))}` `=(28)/(16)=7/4` |
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