## Calculus 10th Edition

The function is continuous when $a=4.$
$\lim\limits_{x\to a^+}\dfrac{x^2-a^2}{x-a}=\lim\limits_{x\to a^+}\dfrac{(x+a)(x-a)}{(x-a)}$ $=\lim\limits_{x\to a^+}(x+a)=a^++a=2a.$ $\lim\limits_{x\to a^-}\dfrac{x^2-a^2}{x-a}=\lim\limits_{x\to a^-}\dfrac{(x-a)(x+a)}{(x-a)}$ $=\lim\limits_{x\to a^-}(x+a)=a^-+a=2a$ Since $\lim\limits_{x\to a^+}\dfrac{x^2-a^2}{x-a}=\lim\limits_{x\to a^-}\dfrac{x^2-a^2}{x-a}\to\lim\limits_{x\to a}\dfrac{x^2-a^2}{x-a}=2a.$ It is also evident from the given that $g(a)=8.$ For a function to be continuous $\lim\limits_{x\to a}g(x)=g(a)\to$ $2a=8\to a=4.$