Given #epsilon > 0# choose #delta = min{1, epsilon/19}#. Note that #delta > 0#
For every #x# with #0 < abs(x-(-2)) < delta#, we have
#abs(x+2) < 1#, so #-1 < x+2 < 1# and #-3 < x < -1#.
This entails that #1 < x^2 < 9# and #2 < -2x < 6#, so that
#7 < x^2-2x+4 < 19#.
Summarizing, for all #x# such that #0 < abs(x-(-2)) < delta#, we have #abs(x^2-2x+4) < 19#.
Now #abs(x^3-(-8)) = abs(x^3+8) = abs(x+2)abs(x^2-2x+4)#
And if #abs(x+2) < delta# then #abs(x+2) < epsilon/19# and #abs(x^2-2x+4) < 19#, so
So, if #abs(x+2) < delta#, then
#abs(x^3-(-8)) = abs(x+2)abs(x^2-2x+4)#
# < (epsilon/19)*(19) = epsilon#.
We have shown that for any positive #epsilon# there is a positive #delta# such that for any #x# with #0 < abs(x-(-2)) < delta#, we have #abs(x^3-(-8)) < epsilon#.
By the definition of limit, #lim_(xrarr-2)x^3 = -8#.
