Calculus 10th Edition

$\lim\limits_{x\to6}(x-2)^2=16.$
By Theorem $1.5: \lim\limits_{x\to c}(f(g(x)))=f(\lim\limits_{x\to c}g(x)).$ $\lim\limits_{x\to6}(x-2)^2=(\lim\limits_{x\to6}(x-2))^2=(6-2)^2=16.$