The first thing you have to do is to #color (red) (plug)# in the value of #x# to test if it doesn't give you an indeterminate form. #0/0# and #oo/oo# are some form of indetermination. I think they are the most popular but we have others as well. If you find some indeterminate form, you will have to use some formulas or tricks but if not, then you are lucky and your answer will be the value you just found.
For this one I can already see that you won't get an indeterminate form because I mentally plugged the value of #x# in the denominator, so I'll spare you some time and not talk about the indeterminate form. You can still ask about it if you want to :)
#lim_(x->2)(x^3-6x-2)/(x^3-4)#
So, plugging #2# in #(x^3-6x-2)/(x^3-4)=((2)^3-6(2)-2)/((2)^3-4)=-3/2#
Hope this helps :)
