How to solve $\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}$ Solve the following question :

\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}\\
\end{eqnarray}

The answer should be $\frac{1}{128}$.
I try that:

\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2} &=& \lim_{x\to 16} \frac{(4-\sqrt{x})(4+\sqrt{x})}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\\ &=& \lim_{x\to 16} \frac{16-x}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\end{eqnarray}

What can I do?

Thank you for your attention.

Hint.
$$(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})=\sqrt{x}(4-\sqrt{x})\cdot \sqrt{x}(4+\sqrt{x})^2=x(16-x)(4+\sqrt{x})$$
