Answer
L = 4, 𝛿 = 0.05
Work Step by Step
L=$\lim\limits_{x \to 2} \frac{x^{2} -4}{x-2}$ = $\lim\limits_{x \to 2}\frac{(x-2)(x+2)}{x-2}$
=$\lim\limits_{x \to 2} x+2$ = 2+2 =4
Solve the inequality |f(x) -L| < e
In the above calculation we concluded = $ \frac{x^{2} -4}{x-2}$ = x+2
|x+2 - 4| <0.05
|x-2| <0.05
-0.050;
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 (1.95, 2.05). distance between 2 and 1.95 is 0.05 and distance between 2 and 2.05 is 0.05. so we will take the smaller value for 𝛿, so 𝛿 = 0.05
