Thomas' Calculus 13th Edition

Solve the inequality |f(x) -L| < e |$\frac{120}{x}$ -5| <1 -1 0; Now we need to find the value of 𝛿>0 that places the open interval (24-𝛿, 24+𝛿) centered at 24 inside the interval (20,30). distance between 24 and 20 is 4 and distance between 24 and 30 is 6. so we will take the smaller value for 𝛿, so 𝛿 =4