<s>[INST] 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.
 [/INST] Putting $\sqrt x=h,$
We have $$\lim_{h\to4}\frac{4-h}{16h^2-h^4}=\lim_{h\to4}\frac{4-h}{h^2(16-h^2)}=\lim_{h\to4}\frac{4-h}{h^2(4-h)(4+h)}=\lim_{h\to4}\frac1{h^2(4+h)}=\cdots$$  Cancelling  out $4-h$ as $4-h\ne0$ as $h\to4$
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