Limit with square root $$\lim_{x\to16}\frac{4-\sqrt{x}}{16x-x^2}$$
I am not sure what to do, I have tried factoring everything and using both conjugates, neither options gives me anything usable. 
 A: First pull out the obvious factor of $x$ in the denominator to get $$\frac{4-\sqrt x}{x(16-x)}\;.$$ The $x$ in the denominator won’t cause any problems in taking the limit, so focus on the rest: $$\frac{4-\sqrt x}{16-x}\;.$$ Notice that each term in the denominator is the square of the corresponding term in the numerator: $$\frac{4-\sqrt x}{16-x}=\frac{4-\sqrt x}{4^2-(\sqrt{x})^2}\;.$$ That last denominator is the difference of two squares; what do you know about factoring such differences?
A: Let's try to get things to look the same:
$$
{4-\sqrt x\over 16x-x^2}={4(1-{\sqrt x\over4})\over 16x(1-{x\over 16}  ) }=
{  1-{\sqrt x\over4} \over 4x(1-{x\over 16} )  }.
$$
Observe that $({\sqrt x\over 4})^2={x\over 16}$ for $x>0$. 
Can you see how to take advantage of the formula $a^2-b^2=(a+b)(a-b)$?
A: Alternative Method: (Assuming one knows differentiation)
$$\lim_{x\to16}\dfrac{4-\sqrt{x}}{16x-x^2}=\lim_{x\to16}\frac1x\dfrac{4-\sqrt{x}}{16-x}=\lim_{x\to16}\frac1x\dfrac{\sqrt{16}-\sqrt{x}}{16-x}.$$
By definition, $\lim\limits_{x\to16}\frac{4-\sqrt{x}}{16-x}$ is the derivative of $\sqrt{x}$ evaluated at $x=16$. And since the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, it follows that its value is: $\frac{1}{2\sqrt{16}}=\frac{1}{8}$. Thus $\lim\limits_{x\to16}\frac{4-\sqrt{x}}{16-x}=\tfrac18$, hence the value of the limit is $\tfrac1{16}\cdot\tfrac18=\tfrac{1}{128}$.
A: Try using L'Hôpital's rule in the case of an indeterminate form.
A: One way is by a rationalizing substitution:
$$
\begin{align}
u & = \sqrt{x} \\  \\
u^2 & = x \\  \\
\text{As }x\to16, & u \to 4.
\end{align}
$$
So
$$\lim_{x\to16}\frac{4-\sqrt{x}}{16x-x^2} = \lim_{u\to4}\frac{4-u}{16u^2-u^4} = \lim_{u\to 4} \frac{4-u}{u^2(4-u)(4+u)} = \lim_{u\to4} \frac{1}{u^2(4+u)}=\frac{1}{4^2(4+4)}=\frac{1}{128}.$$
