# 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.

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You can always use L'Hôpital's Rule – user14783 Aug 19 '11 at 5:56
This should probably have been a comment, but you cannot yet make them. – Mariano Suárez-Alvarez Aug 19 '11 at 5:57
...but if you can, avoid using it. :) – J. M. Aug 19 '11 at 6:52
@Ronnie: I have converted your answer to a comment. As Mariano mentioned, because you do not have 50 reputation points yet, you can only comment on your own questions and answers. So, you didn't do anything wrong; the "add comment" button will only appear for you once you gain 50 points. Here is an explanation of reputation points. – Zev Chonoles Aug 19 '11 at 14:50

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.

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I tried this, but ended up with the same exact problem I started with. – Joe Aug 19 '11 at 5:34
@Joe: What did you get after performing this multiplication? Did you cancel what can be cancelled? Alternatively: $x-16=(\sqrt x-4)(\sqrt x+4)$ – J. M. Aug 19 '11 at 5:37
I hadn't noticed you could cancel x-16 with 16-x. The answer my teacher gave us was -8, though. How is that? – Joe Aug 19 '11 at 5:44
Is it negative because it's 16-x and x-16? Would x-16 and x-16 make it positive? I'm sort of confused with the signs here. – Joe Aug 19 '11 at 5:56
@Joe: Yes, $x-16=(-1)(16-x)$, so when you divide $x-16$ by $16-x$, you get $-1$. And yes, $(x-16)/(x-16)=1$, positive. – André Nicolas Aug 19 '11 at 6:09

$$\frac{{x - 16}}{{4 - \sqrt x }} = \frac{{(x - 16)(4 + \sqrt x )}}{{(4 - \sqrt x )(4 + \sqrt x )}} = \frac{{(x - 16)(4 + \sqrt x )}}{{16 - x}} = - (4 + \sqrt x ) \to - 8$$ as $x \to 16$.

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