How do you prove that the limit #(x^2+x-4)=8# as x approaches 3 using the formal definition of a limit?

1 Answer
Jan 1, 2017

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}#