Let #f(x)=(sin(5x)cot(3x))/(xcot(4x))#. Then
#f(x)=(60x)/(60x)*f(x)=(60xsin(5x)cot(3x))/(60x^{2}cot(4x))#
#=(60xsin(5x)cos(3x)sin(4x))/(60x^{2}sin(3x)cos(4x))#
#=(3x)/(sin(3x))* sin(5x)/(5x) * sin(4x)/(4x) * (20cos(3x))/(3cos(4x))#
Since #lim_{x->0}sin(x)/x=1#, it follows that #lim_{x->0}(3x)/(sin(3x))=1#, #lim_{x->0}sin(5x)/(5x)=1#, and #lim_{x->0}sin(4x)/(4x)=1#.
Moreover, we can say that #lim_{x->0}(20cos(3x))/(3cos(4x))=(20 * 1)/(3 *1) = 20/3# by substitution into the function since #(20cos(3x))/(3cos(4x))# is continuous at #x=0#.
Finally, applying the rule for the limit of a product of functions (whose individual limits exist), we obtain
#lim_{x->0}f(x)=1 * 1 * 1 * 20/3 = 20/3#.
