The product of all possible values of α, for which
\[\lim_{x\to0}\left(\frac{1-\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right)=2\]
is:
Solutions · 1
Official solution
Using the standard small-angle result 1−cosAcosBcosC≈2A2+B2+C2 (for A,B,C→0), with A=αx,B=(α+1)x,C=(α+2)x, and sin2((α+1)x)≈(α+1)2x2:
\[\lim_{x\to0}\frac{\frac12[\alpha^2+(\alpha+1)^2+(\alpha+2)^2]x^2}{(\alpha+1)^2x^2}=2\]
\[\alpha^2+(\alpha+1)^2+(\alpha+2)^2=4(\alpha+1)^2 \implies \alpha^2+(\alpha+2)^2=3(\alpha+1)^2\]
Expanding: 2α2+4α+4=3α2+6α+3⟹α2+2α−1=0
By Vieta's formulas, the product of the roots is 1−1=−1.