Question 1069293
{{{-1<=cos(12pi*x)<=1}}}
{{{-x^2<=x^2cos(12pi*x)<=x^2}}}
So then on the left hand side,
{{{lim(x->0,(-x^2))=0}}}
and on the right hand side,
{{{lim(x->0,(x^2))=0}}}
So by the squeeze theorem,
{{{lim(x->0,(x^2*cos(12pi*x)))=0}}}