## Thomas' Calculus 13th Edition

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