# "First, let's recall a key, and fundamental, trig limit. We have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad lim_{A rarr 0} sinA/A \ = \ 1. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (I) #
# "Now, let's look at the limit in question. We see:" #
# \qquad \qquad \qquad lim_{x rarr 0} ( 3 x^4 )/sin^2( 3 x^2 ) \ = \ lim_{x rarr 0} 3/3 cdot ( 3 x^4 )/sin^2( 3 x^2 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ lim_{x rarr 0} 1/3 cdot ( 9 x^4 )/sin^2( 3 x^2 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ lim_{x rarr 0} 1/3 cdot ( 3 x^2 )^2/( sin( 3 x^2 ) )^2 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ lim_{x rarr 0} 1/3 cdot ( ( 3 x^2 )/( sin( 3 x^2 ) ) )^2 #
# \qquad \quad "using eqn." \ (I) \ "above [and noting that, when" \ x rarr 0, #
# \qquad \quad "we also have" \ \ 3 x^2 rarr 0], "we continue calculating:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = 1/3. #
# "This is our answer !!" #
# "So, summarizing, we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ lim_{x rarr 0} ( 3 x^4 )/sin^2( 3 x^2 ) \ = \ 1/3. #
