We want to make #abs((x^2+x-4)-8) < epsilon#
We control #delta# and #abs(x-3) < delta#
Note that:
#abs((x^2+x-4)-8) =abs(x^2+x-12)#
#= abs((x+4)(x-3))#
# = abs(x+4) abs(x-3)#
We control #abs(x-3)# and so, indirectly, we control #abs(x+4)#
Make #abs(x-3) < 1# . (Any positive number will work in place of #1#, but the details of what follows will change.)
This assures us that #2 < x < 4#.
So considering #x+4# ( the other factor), we see:
#6 < x+4 < 7.# And
#abs(x+4) < 7#.
So, if #abs(x-3) < 1#, then
#abs((x^2+x-4)-8) = abs(x+4) abs(x-3) < 7abs(x-3)#
If we also make sure that #abs(x-3) < epsilon/7#, then we will have:
#abs((x^2+x-4)-8) < 7abs(x-3) < 7(epsilon/7) = epsilon#.
Now write the proof :
Given #epsilon > 0#, let #delta =min {1, epsilon/7}#.
Now is #abs(x-3) < delta#, we will have:
#abs((x^2+x-4)-8) = abs(x^2+x-12) = abs(x+4) abs(x-3)#
# < 7 (epsilon/7) = epsilon#
