Answer
$\lim\limits_{x \to 16}\frac{4-\sqrt x}{16x-x^{2}}=\frac{1}{128}$
Work Step by Step
$\lim\limits_{x \to 16}\frac{4-\sqrt x}{16x-x^{2}}=\lim\limits_{x \to 16}\frac{4-\sqrt x}{x(16-x)}=\lim\limits_{x \to 16}\frac{4-\sqrt x}{x(4-\sqrt x)(4+\sqrt x)}=\lim\limits_{x \to 16}\frac{1}{x(4+\sqrt x)}=\frac{1}{16\times(4+4)}=\frac{1}{128}$
