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#