Notice that plugging in #x=-2# in the function's current form gives #(-2+2)/((-2)^2+6(-2)+8)=0/0#, so the limit is currently in an indeterminate form and we cant yet determine the limit.
However, notice that we can factor the denominator of the fraction:
#lim_(xrarr-2)(x+2)/(x^2+6x+8)=lim_(xrarr-2)(x+2)/((x+2)(x+4))#
Now we can see why plugging in #x=-2# into the denominator creates a #0#. However, we can cancel the #(x+2)# terms in the numerator and denominator:
#=lim_(xrarr-2)1/(x+4)#
In other words, the function #(x+2)/(x^2+6x+8)# is identical to the function #1/(x+4)# except at the point #x=-2#. The former function is identical to the latter function except that the first function has a hole at #x=-2#. So, the limit is where the function should lie, which can be found by plugging #x=-2# into the simplified function, #1/(x+4)#.
#=1/(-2+4)=1/2#
