# Limits of an Absolute Value

1. Oct 10, 2011

### smerhej

Find the limit, if it exists. (If an answer does not exist, enter DNE.)
lim x→−8 [ 8 − |x| OVER 8 + x ]

The attempt at a solution

First → abs(x) = -x ; x<0 / x ; x≥0

Therefore, our equation is going to look like 8 − -x OVER 8 + x

If approaching -8 from the left, I got 8 − -(-8-) OVER 8 + -8-

→ 0- OVER 0- . When dividing 0- over 0- you get infinity, correct?

And doing the limit as x approaches -8 from the right gave me infinity as well, thus giving the answer, that as X approaches -8, Y approaches infinity. But that's wrong. Can someone help please?

2. Oct 10, 2011

### ehild

What is 8-(-x)???? Is not it the same as the denominator?

ehild

3. Oct 10, 2011

### SammyS

Staff Emeritus
0 over 0 is not infinity. It's an indeterminate form.

(8-(-x))/(8+x) = ? provided that x ≠ -8
(parentheses are important).

4. Oct 10, 2011

### HallsofIvy

Staff Emeritus
The crucial point is that, for x going to -8, you can assume that x< 0. |x|= -x.

5. Oct 10, 2011

### smerhej

Right, so I get the equation (8+x)/(8+x) = ? . Now I'm not entirely sure where to go from here.

Just simply putting the value -8 into the equation gives 0/0 (which is wrong), and I'm not entirely sure how to change the way this equation looks..

Would saying that it equals 1 be fair? Seeing as how the numerator and the denominator are the same?

6. Oct 10, 2011

### Staff: Mentor

Fair has nothing to do with it. Since the numerator and denominator are the same, for all negative values of x other than -8, the value of the expression (8 + x)/(8 + x) is 1. From this, you should be able to say what the value of the limit is.

7. Oct 10, 2011

### ehild

There is a definition, that a function f(x) has the limit A at x0 if to every sequence xn convergent to x0 the sequence f(xn) converges to A. This way a function can have a limit where it is not defined. Such an example is limx-->1f(x)=(x^2-1)/(x-1). f(x) is not defined at x=1, but for all x≠1 it is equal to x+1, so its limit is 2.

ehild

8. Oct 11, 2011

### smerhej

Ah thank you very much! And you are right, fair does have nothing to do with it.. I'll be sure to be more careful with my words.