Does the Limit Exist for Absolute Value as x Approaches -8?

[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?

 

[ehild]

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

ehild

 

[SammyS]

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

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

 

[HallsofIvy]

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

 

[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?

 

[Mark44]

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?

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.

 

[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

 

[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.