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.
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 $h\to4$
Hint.
$$(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})=\sqrt{x}(4-\sqrt{x})\cdot \sqrt{x}(4+\sqrt{x})^2=x(16-x)(4+\sqrt{x})$$
I think you can do it the way you started, as follows:
$$\frac {4-\sqrt x}{16x-x^2}\cdot \frac {4+\sqrt x}{4+\sqrt x}=\frac {16-x}{x(16-x)(4+\sqrt x)}=\frac 1{x(4+\sqrt x)}$$
You made it too complicated by factoring the bottom (denominator) the way you did. With the limit at $16$, look to cancel a factor $(x-16)$. You can also cancel $4-\sqrt x$ directly and obtain the same result.
