We want to find $\lim_{x\to 16}\frac{x-16}{4-\sqrt{x}}.$
Multiply "top" and "bottom" by $4+\sqrt{x}$. That is perfectly legitimate, we are multiplying our expression by $1$. Since $(4-\sqrt{x})(4+\sqrt{x})=16-x$, $\lim_{x\to 16}\frac{x-16}{4-\sqrt{x}}=\lim_{x\to 16}\frac{(x-16)(4+\sqrt{x})}{(4-\sqrt{x})(4+\sqrt{x})}=\lim_{x\to 16}\frac{(x-16)(4+\sqrt{x})}{16-x}.$
When $x \ne 16$, our expression simplifies to $-(4+\sqrt{x})$. This is because $x-16=(-1)(16-x)$. So our limit is equal to $\lim_{x\to 16} -(4+\sqrt{x}).$
It is clear that this last limit is $-8$. If we want to mention fine details, $\displaystyle\lim_{x\to 16}\sqrt{x}=4$ because $\sqrt{x}$ is continuous at $x=16$.
Remark: Here is an alternative way of doing the same thing. Note that for non-negative $x$, $x-16=(\sqrt{x}-4)(\sqrt{x}+4)$.
Thus we are interested in $\lim_{x\to 16}\frac{(\sqrt{x}+4)(\sqrt{x}-4)}{4-\sqrt{x}}.$ The rest is straightforward.