Solving for a limit with an indeterminate fraction and square root of X I have a test tomorrow and there's one thing I found that I don't understand while reviewing. I've looked all over YouTube and Google and just can't find out how to work this problem:
$$ \lim_{x\to16} \frac{x-16}{4-\sqrt{x}} $$
It's obvious that it's indeterminate but I can't get any further than that.
 A: 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. 
A: $$
\frac{{x - 16}}{{4 - \sqrt x }} = \frac{{(x - 16)(4 + \sqrt x )}}{{(4 - \sqrt x )(4 + \sqrt x )}} = \frac{{(x - 16)(4 + \sqrt x )}}{{16 - x}} =  - (4 + \sqrt x ) \to  - 8
$$
as $x \to 16$.
