Given #epsilon > 0# choose #delta = min{1, epsilon/8}#. Note that #delta > 0#.
For every #x# with #0 < abs(x-3) < delta#, we have
#abs(x-3) < 1#, so #-1 < x-3 < 1# and #2 < x < 4#.
This entails that #6 < x+4 < 8# , so that #abs(x+4) < 8#.
Summarizing, for all #x# such that #0 < abs(x-3) < delta#, we have #abs(x+4) < 8#.
Now #abs((x^2+x-4)-8) = abs(x^2+x-12) = abs(x+4)abs(x-3)#
And if #abs(x-3) < delta# then #abs(x+4) < 8#, and #abs(x-3) < epsilon/8#.
So, if #abs(x-3) < delta#, then
#abs((x^2+x-4)-8) = abs(x+4)abs(x-3)#
# < (8)*(epsilon/8) = epsilon#.
We have shown that for any positive #epsilon# there is a positive #delta# such that for any #x# with #0 < abs(x-3) < delta#, we have #abs((x^2+x-4)-(8)) < epsilon#.
By the definition of limit, #lim_(xrarr-2)(x^2+x-4)= 8#.
Additional Note
The choice to use #1# as one of the bounds on #delta# was arbitrary. We could have chosen another (positive) value, but once chosen, that would determine the second bound on #delta#.
For example, if we had used #2#, then we would have gotten:
For every #x# with #0 < abs(x-3) < delta#, we have
#abs(x-3) < 2#, so #-2 < x-3 < 2# and #1 < x < 5#.
This entails that #5 < x+4 < 9# , so that #abs(x+4) < 9#.
Summarizing, for all #x# such that #0 < abs(x-3) < delta#, we have #abs(x+4) < 9#.
We would have used #delta = min{2,epsilon/9}#
