Answer
L=4, 𝛿 = 0.75
Work Step by Step
L=$\lim\limits_{x \to -3} \sqrt 1-5x$
=$\lim\limits_{x \to -3}\sqrt 1-5.(-3)$ =$\sqrt 1+15$ = 4
Solve the inequality |f(x) -L| < e
|$\sqrt 1-5x$ - 4 | <0.5
-0.50;
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 (-3.85, -2.25). distance between -3 and -3.85 is 0.85 and distance between -3 and -2.25 is 0.75 . so we will take the smaller value for 𝛿, so 𝛿 = 0.75
