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Math Help - absolute value in limits

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by haebinpark View Post
    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?
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  3. #3
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    Oct 2009
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    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 ?
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