I need help with this problem please. I did it but it was false, I attached the feedback and the example of 3.7.5.

Consider the function f(x) = ^3 for x ? R and -2

20. Consider the function f(x) = x for x ER and -2 1 f(x) = 1. Use the method shown in Example 3.7.5.Approach to Solution Our goal is to first find a positive number & We have: e>0 8 = some positive number 00 Approach to Solution 4 |x3 - 1| = |(x - 1)(x2 + x + 1)| = (x2 + x + 1)|x - 1| |x - 1| 0 then & is a positive number. |x3 - 1| = |(x - 1)(x2 + x + 1)| = (x2 + x+ 1)|x - 1| 0 Let delta = min{1, epsilon/7} Let x be a real number such that -2 |x-1]-101 (x"2 + x + 1)(delta) [f(x) - 1| 035>0Vx(0-::|x3]';7' \\'7 x3 % e 2x 5x 3_71{6)' Vx({){|x3|{5> x-3 Before working out the value of , let's figure out what the rest of the proof will look like. Based on the form of the goal at this point, we should proceed by letting x be arbitrary, assuming 0 OVx(0 038 > OVx(0 0, this is a positive number, as required. Solution 2t 55 Theorem. lim = >3 _1q, x3 x3 Proof. Suppose > 0. Let = /2, which is also clearly positive. Let x be an arbitrary real number, and suppose that 0 0, this is a positive number, as required. Solution 2x2 -5x - 3 Theorem. lim = 7. x-+ 3 x - 3 Proof. Suppose E > 0. Let 8 = E/2, which is also clearly positive. Let x be an arbitrary real number, and suppose that 0