# \ #
# "We compute, and use L'Hosptal's Rule:" #
# lim_{x rarr 0} { 1 - cos(x)cos(2x)cos(3x) }/{Sin^2(2x)} \qquad \qquad 0/0, "use L'Hospital's Rule" #
# \qquad \qquad = \ lim_{x rarr 0} { [ 1 - cos(x) cos(2x)cos(3x) ]' }/{ [ Sin^2(2x) ]' } #
# \ #
# \qquad \qquad = \ lim_{x rarr 0} - { [ cos(x) ]'cos(2x)cos(3x) + cos(x)[ cos(2x) ]'cos(3x) + cos(x)cos(2x)[ cos(3x) ] '}/{ 2 \cdot sin(2x) \cdot 2 } #
# \ #
# \qquad \qquad = \ lim_{x rarr 0} - { [- sin(x) ]cos(2x)cos(3x) + cos(x)[ -2 sin(2x) ]cos(3x) + cos(x)cos(2x)[ -3sin(3x) ] }/{ 2 \cdot sin(2x) \cdot 2 } #
# \ #
# \qquad \qquad = \ lim_{x rarr 0} { sin(x)cos(2x)cos(3x) + 2 cos(x)sin(2x)cos(3x) + 3 cos(x)cos(2x)sin(3x) }/{ 2 \cdot sin(2x) \cdot 2 } #
# \ #
# \qquad \qquad = \ lim_{x rarr 0} { sin(x)cos(2x)cos(3x) + 2 cos(x)( 2 sin(x)cos(x))cos(3x) + 3 cos(x)cos(2x)( 3 cos^2(x)sin(x) - sin^3(x) ) }/{ 4 \cdot ( 2sin(x) cos(x) ) } #
# \ #
# "[Remove factor of" \ sin(x) \ "from top and bottom:]"#
# \qquad \qquad = \ lim_{x rarr 0} { cos(2x)cos(3x) + 2 cos(x)( 2 cos(x))cos(3x) + 3 cos(x)cos(2x)( 3 cos^2(x) - sin^2(x) ) }/{ 4 \cdot ( 2 cos(x) ) } #
# \ #
# "[Substitute" \ x=0 ":]" #
# \qquad \qquad = \ { 1 \cdot 1 + 2 \cdot 1( 2 \cdot 1) \cdot 1 + 3 \cdot 1 \cdot 1( 3 \cdot 1^2 - 0^2 ) }/{ 4 \cdot ( 2 \cdot 1 ) } #
# \qquad \qquad = \ { 1 + 4 + 9 }/{ 8 } \ = \ {14}/{ 8 } #
# \qquad \qquad = 7/4.#
# \ #
# "So, following the statement of the problem, the answer is:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad "Choice C." \quad \quad \quad \ square #
