The function #f(x)=arctan(x^2-4)/(3x^2-6x)=arctan(x^2-4)/(3x(x-2))# is known to be continuous everywhere except #x=0# and #x=2# (this could be proved, but it would be a lot of work).
To "evaluate a limit by using continuity" means that you can evaluate the limit of a known continuous function at a point of continuity just by substituting the point into the function. That is, if #f# is defined near #x=a# and continuous at #x=a#, then #lim_{x->a}f(x)=f(a)#. Applying this fact to the function #f# gives:
#lim_{x->a}f(x)=lim_{x->a}arctan(x^2-4)/(3x^2-6x)=f(a)=arctan(a^2-4)/(3a^2-6a)#
when #a!=0# and #a!=2#.
This is not part of the given question, but it turns out that, even though #f(2)# does not exist and #f# is not continuous at #x=2#, the limit #lim_{x->2}f(x)# does exist in this example. In fact, from L'Hopital's Rule , it can be shown that #lim_{x->2}arctan(x^2-4)/(3x^2-6x)=2/3#. Try seeing if you can prove this!
