## Thomas' Calculus 13th Edition

L=$\lim\limits_{x \to 3}$ 3-2x = 3 -2.3 = 3 - 6 = -3 Solve the inequality |f(x) -L| < e |3-2x+3| <0.02 |6-2x| <0.02 -0.02<6-2x<0.02 -6.02\$0; Now we need to find the value of 𝛿>0 that places the open interval (3-𝛿, 3+𝛿) centered at 3 inside the interval (2.99, 3.01). distance between 3 and 2.99 is 0.01 and distance between 3 and 3.01 is 0.01. so we will take the smaller value for 𝛿, so 𝛿 = 0.01