4
$\begingroup$

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.

  • 0
    @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](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757). 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](http://meta.stackexchange.com/questions/7237/how-does-reputation-work/7238#7238).2011-08-19

2 Answers 2

4

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.

  • 0
    @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.2011-08-19
1

$ \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 $x \to 16$.