## Thomas' Calculus 13th Edition

($\frac{-0.03}{m}$+2, $\frac{0.03}{m}$+2) 𝛿 =$\frac{0.03}{m}$
Solve the inequality |f(x) -L| < e |mx-2m| <0.03 |m(x-2)| <0.03 -0.03< 0; Now we need to find the value of 𝛿>0 that places the open interval (2-𝛿, 2+𝛿) centered at 2 inside the interval ($\frac{-0.03}{m}$+2, $\frac{0.03}{m}$+2). distance between 2 and $\frac{-0.03}{m}$+2 is $\frac{0.03}{m}$ and distance between 2 and $\frac{0.03}{m}$ is $\frac{0.03}{m}$. so we will take the smaller value for 𝛿, so 𝛿 =$\frac{0.03}{m}$