Question

Limx---0 ( 1- cosx cos2x cos3x) /(sin 2 2x) = ?

Answer 1

Answer:

7/4

Step-by-step explanation:

$lim\:\frac{1-cosx\:cos2x\:cos3x}{sin^22x}\\ x\rightarrow 0\\=lim\:\frac{1-cosx\:cos2x\:cos3x}{1-cos^22x}\\ x\rightarrow 0\\=lim\:\frac{1-\frac{cos2x}{2}[cos4x+cos2x]}{1-cos^22x} \\ x\rightarrow 0\\=lim\:\frac{1-\frac{cos2x}{2}[2cos^22x-1+cos2x]}{1-cos^22x} \\ x\rightarrow 0$

let cos2x = t

$lim\:\frac{2t^3+t^2-t-2}{2(t^2-1)}\\t\rightarrow 1\\lim\:\frac{2t^3-2+t^2-t}{2(t+1)(t-1)}\\t\rightarrow 1\\lim\:\frac{2(t-1)(t^2+t+1)+t(t-1)}{2(t+1)(t-1)}\\t\rightarrow 1\\=\frac{2*3+1}{2*2}\\=\frac{7}{4}$