How do you prove that the limit of #(x^2) =9 # as x approaches -3 using the epsilon delta proof?

Question

Scarlett Allison · Accepted Answer

See explanation...

Let #epsilon > 0# Choose #delta in (0, epsilon / 7) nn (0, 1)# Note that: #0 < delta^2 < delta# If #x in (-3-delta, -3+delta)# then: #abs(x^2 - 9) < abs((-3-delta)^2 - 9) = abs((-3)^2+6delta+delta^2-9) < 7delta < epsilon# So: #AA epsilon > 0 EE delta > 0 : AA x in (-3-delta, -3+delta), abs(x^2-9) < epsilon# So: #lim_(x->-3) x^2 = 9#

Emilia Aguilar · Answer

undefined To prove that the limit of (x^2) as x approaches -3 is 9 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - (-3)| < delta, then |(x^2) - 9| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a suitable delta > 0. We start by considering the expression |(x^2) - 9| and try to manipulate it to fit our requirements. |(x^2) - 9| = |x + 3||x - 3| Now, we can see that if |x + 3| < 1, then |(x^2) - 9| = |x + 3||x - 3| < (1)(|x - 3|). So, we can choose delta = 1. Now, let's assume that 0 < |x - (-3)| < delta = 1. |x - (-3)| = |x + 3| < 1. From this, we can deduce that -4 < x < -2. Now, let's consider |(x^2) - 9| again: |(x^2) - 9| = |x + 3||x - 3| < (1)(|x - 3|). Since -4 < x < -2, we can conclude that -1 < x - 3 < 1. Therefore, |(x^2) - 9| < (1)(|x - 3|) < 1. Hence, we have shown that for any epsilon > 0, we can choose delta = 1, and if 0 < |x - (-3)| < delta, then |(x^2) - 9| < epsilon. Therefore, the limit of (x^2) as x approaches -3 is indeed 9, as proven using the epsilon-delta proof.