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Math Help - finding limits

  1. #1
    Junior Member
    Joined
    Mar 2009
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    32

    finding limits

    having lots of trouble finding limits

    1:
    lim as v -> -2 from left of {v-2 / |v-2|}

    2:
    lim as v -> -2 from right of {v-2 / |v-2|}

    for these 2, i'm thinking that |v-2| is 2 - v if v <= 2 and v - 2 if x > 2

    i got that far but don't know what to do next

    3:
    lim as x -> 1 of cos^-1 {1 - sqrt(x) / 1 - 2}

    the cos^-1 is what has me stuck, not sure how to interpret

    4:
    lim as x-> infinity of {sqrt(x^2 -9) / 2x - 6}

    what i did was multiply top/bottom by sqrt(x^2 -9) to make numerator x^2 - 9 but then the not sure what to do w/ the denominator

    5:
    lim as x -> infinity of {x^4 + 2 / (x^2 - 1) (2x^2 + 1)

    this one i tried to factor out the numerator but it doesn't factor out properly for it to cancel out the denominator's (x^2 - 1). unless i factored wrong, what am i missing/doing wrong with this one?
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  2. #2
    DBA
    DBA is offline
    Member
    Joined
    May 2009
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    129
    For 1 and 2:
    is it the limit as v -> -2 (negative 2)?

    |v-2| => (v-2) for v>= 2 and
    -(v-2) for v< 2

    In both cases you described, x would be smaller than 2, Since -2< 2

    Then both limits would be the same

    from the left: lim v-> -2- v-2/|v-2| = (v-2)/-(v-2) = -1
    from the right: lim v-> -2+ v-2/|v-2| = (v-2)/-(v-2) = -1
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