# Limits: Zero / Non-Zero

1. Oct 16, 2008

### Alexstre

Hello!

I'm trying to find the following limit:
$${\lim_{x \to 5}}\ {5-x \over {3-\sqrt{x^2 -16}}$$

I tried 2 things
Simplifying the bottom:
$$3-\sqrt{x^2-16}$$ = $$3-\sqrt{x^2-4^2}$$ = 3-x-4 = -1-x

But that doesn't help with what's on top...

I also tried multiplying top and bottom by:
$$3+\sqrt{x^2-16}$$ but I still ended up with 0/non-zero.

Could anyone point me into the right direction, or correct me if one of those 2 steps was right?

Thanks!

2. Oct 16, 2008

### EnumaElish

How do you figure Sqrt(x^2 - 4^2) = x - 4?

Have you thought of differentiating the numerator and the denominator, then taking the limit of their ratio?

3. Oct 16, 2008

### MATdaveLACK

If the limit is zero over non-zero, the limit is just zero. But in this case I'm getting the limit to be 0/0, yeah?

4. Oct 17, 2008

### HallsofIvy

Staff Emeritus
This is wrong. $\sqrt{x^2- 4^2}$ is NOT x- 4.

Good! That prevented you from getting the wrong answer!

Really? WHAT did you get when you did that? I do NOT get "0/non-zero".

I recommend you recheck your algebra. In particular, what is
$$\left(\frac{5-x}{3-\sqrt{x^2- 16}}\right)\left(\frac{3+\sqrt{x^2- 16}}{3+\sqrt{x^2- 16}}\right)$$

By the way, as MATdaveLACK said, there is nothing wrong with getting "0/non-zero": that would just mean the limit is 0. However, here, that is wrong.