Answer
($\frac{-3}{m}$+3, $\frac{3}{m}$+3), 𝛿 =$\frac{3}{m}$
Work Step by Step
Solve the inequality |f(x) -L| < e
|mx-3m| <3
|m(x-3)| <3
-3< 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($\frac{-3}{m}$+3, $\frac{3}{m}$+3). distance between 3 and $\frac{-3}{m}$+3 is$\frac{3}{m}$ and distance between 3 and $\frac{3}{m}$+3 is $\frac{3}{m}$. so we will take the smaller value for 𝛿, so 𝛿 =$\frac{3}{m}$
