# Thread: absolute value in limits

1. ## absolute value in limits

the question is
use the definition of continuity to determine if f(x) is continuous at x = 0
if not, give type of discontinuity

f(x) = abs (x+3) x < 0
2x + 1 x = 0
(x+2)^2 - 1 x > 0

the problem im having with is "abs (x+3) x < 0" this part
as far as i know, since x < 0, x is approaching from the left
thus, abs (x+3) = -(x+3) ??????
but 0+3 > 0...? so +(x+3)?????
im really confusing if x+3 has to be negative or positive

someone put me in the right track please !!

thank you

2. Originally Posted by haebinpark
the question is
use the definition of continuity to determine if f(x) is continuous at x = 0
if not, give type of discontinuity

f(x) = abs (x+3) x < 0
2x + 1 x = 0
(x+2)^2 - 1 x > 0

the problem im having with is "abs (x+3) x < 0" this part
as far as i know, since x < 0, x is approaching from the left
thus, abs (x+3) = -(x+3) ??????
but 0+3 > 0...? so +(x+3)?????
im really confusing if x+3 has to be negative or positive

someone put me in the right track please !!

thank you
$x+3> 0$ if $x> -3$. So $|x+3|= x+3$ as long as $x\ge -3$. And, of course, since the problem is really "at" x= 0, if x is close to 0, it is larger than -3, whether x itself is negative or not.

As far as the limit at x= 0 is concerned, you can treat this function as just x+ 3 for x< 0, $(x+2)^2- 1$ for x> 0. What are the two one-sided limits? Does the limit itself exist? Is that limit equal to the value of the function at x=0?

3. i don't understnad when you said "x+3 > 0 if x>-3" ?
since the question was abs (x+3) x < 0

shouldn't it be x+3 < 0 ? not x+3 > 0 ?