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Math Help - Limits and Continuity

  1. #1
    Newbie
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    Limits and Continuity

    Hey all,

    It seems easy but I dont seem to be getting the answer I am looking for

    Let f(x)=(x+3) if x is less than -2
    (Square root of(x+2) if x is greater than 1

    I am looking to apply these too:

    A) lim f(x) &
    as x approches -2+

    B) Lim f(x)
    as x approaches -2-


    Thanks alot, I am really happy to have found this site and I look forward to being a member.
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  2. #2
    MHF Contributor
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    A) lim f(x) &
    as x approches -2+


    Since in the interval -2 <= x <= 1, or [-2,1], the f(x) is not defined, then as x approaches -2 from the right, it is stopped from moving beyond x=1, where f(x) = f(1) = sqrt(1+2) = sqrt(3).
    Therefore the limit is sqrt(3).

    B) Lim f(x)
    as x approaches -2
    -

    As x approaches -2 from the left, the limit of f(x) is the intersection of f(x) = x+3 and the vertical line x = -2
    So,
    f(x) = x+3, and when x = -2, f(x) = -2 +3 = 1.
    Therefore, the limit is 1.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Svt-Sonic View Post
    Hey all,

    It seems easy but I dont seem to be getting the answer I am looking for

    Let f(x)=(x+3) if x is less than -2
    (Square root of(x+2) if x is greater than 1

    I am looking to apply these too:

    A) lim f(x) &
    as x approches -2+
    This is undefined as f(x) is not defined arbitarily close to -2 from the right.

    B) Lim f(x)
    as x approaches -2-
    this limit is +1, as f(x) is left continuous at -2.

    RonL
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  4. #4
    Newbie
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    Oct 2007
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    Thanks alot guys!
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