Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 43

lim_(x->0) cot(2x)sin(6x)
The function cot(2x)sin(6x) is undefined at x=0. So to take its limit, let's apply the L'Hospital's Rule. To do so, express it as a rational function.
=lim_(x->0) cos(2x)/sin(2x) * sin (6x) = lim_(x->0) (cos(2x)sin(6x))/sin(2x)
Then, take the derivative of the numerator and denominator.
=lim_(x->0) ((cos(2x)sin(6x))')/((sin(2x))')=lim_(x->0) (cos(2x)*cos(6x)*6 + (-sin(2x))*2*sin(6x)) / (cos(2x)*2)
=lim_(x->0) (6cos(2x)cos(6x) - 2sin(2x)sin(6x))/(2cos(2x))=lim_(x->0) (3cos(2x)cos(6x) - sin(2x)sin(6x))/(cos(2x))
And, plug-in x=0.
=(3cos(2*0)cos(6*0)-sin(2*0)sin(6*0))/cos(2*0)=(3*1*1-0)/1=3

Therefore, lim_(x->0) cot(2x)sin(6x) = 3 .