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!
