## Calculus 10th Edition

$\lim\limits_{x\to4}\dfrac{x^2-16}{x-4}=8.$
$\lim\limits_{x\to4}\dfrac{x^2-16}{x-4}=\lim\limits_{x\to4}\dfrac{(x-4)(x+4)}{(x-4)}$ $=\lim\limits_{x\to4}(x+4)=4+4=8.$ N.B: Possible typo in book where the variables are not the same; in this solution $x$ was used.