Answer
L=2, 𝛿 =0.333
Work Step by Step
L = $\lim\limits_{x \to 2} \frac{4}{x}$ = $ \frac{4}{2}$ = 2
Solve the inequality |f(x) -L| < e
|$\frac{4}{x}$ -2| <0.4
-0.4 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 (1.667, 2.5). distance between 1.667 and 2 is 0.333 and distance between 2 and 2.5 is 0.5. so we will take the smaller value for 𝛿, so 𝛿 =0.333
