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

Question

Cameron Alexander · Accepted Answer

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}#

Isabella Ackerman · Answer

undefined To prove that the limit of the function (x^2+x-4) is 8 as x approaches 3 using the formal definition of a limit, we need to show that for any positive value ε, there exists a positive value δ such that if 0 < |x - 3| < δ, then |(x^2+x-4) - 8| < ε. Let's proceed with the proof: Given the function f(x) = x^2 + x - 4, we want to show that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 3| < δ, then |(x^2+x-4) - 8| < ε. First, let's simplify the expression |(x^2+x-4) - 8|: |(x^2+x-4) - 8| = |x^2 + x - 12| Now, we can factor the expression x^2 + x - 12: x^2 + x - 12 = (x - 3)(x + 4) Next, we can rewrite the absolute value expression using the factored form: |(x^2+x-4) - 8| = |(x - 3)(x + 4)| Since we are interested in the behavior of the function as x approaches 3, we can assume that 0 < |x - 3| < 1 (we will use this assumption later). Now, let's consider the expression |(x - 3)(x + 4)|: |(x - 3)(x + 4)| = |x - 3||x + 4| Since we have assumed that 0 < |x - 3| < 1, we can also assume that 0 < x - 3 < 1, which implies -1 < x - 3 < 0. Using this assumption, we can establish a bound for |x + 4|: -1 < x - 3 < 0 3 < x < 4 7 < x + 4 < 8 Therefore, we can conclude that |x + 4| < 8. Now, let's choose δ = min{1, ε/8}. If 0 < |x - 3| < δ, then we have: 0 < |x - 3| < 1 |x - 3||x + 4| < δ|x + 4| < (ε/8) * 8 = ε Hence, we have shown that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 3| < δ, then |(x^2+x-4) - 8| < ε. Therefore, the limit of the function (x^2+x-4) as x approaches 3 is indeed 8, as desired.