Answer
L=1, 𝛿 = 0.01
Work Step by Step
L=$\lim\limits_{x \to -1}$ -3x-2 = -3.(-1) -2= 3 - 2= 1
Solve the inequality |f(x) -L| < e
|-3x-2-1| <0.03
|-3x-3| <0.03
-0.030;
Now we need to find the value of 𝛿>0 that places the open interval
(-1-𝛿, -1+𝛿) centered at -1 inside the interval (-1.01, -0.99). distance between -1 and -1.01 is 0.01 and distance between -1 and -0.99 is 0.01. so we will take the smaller value for 𝛿, so 𝛿 = 0.01
