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Limit x approches 1(sqrt(1+2x)-(sqrt3x))/(sqrt(3+x)-2sqrt(x))

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  1. lim (x->1) (√(1+2x)-√(3x))/(√(3+x)-2√x)

    The fraction at x=1 is 0/0, so we have to try to resolve that. Multiplying by the conjugate of the denominator we have

    (√(1+2x)-√(3x))/(√(3+x)-2√x)) * (√(3+x)+2√x)/(√(3+x)+2√x)

    (√(1+2x)-√(3x))*(√(3+x)+2√x) / (3+x - 4x)

    as x->1 that becomes

    ((√3-√3)(√3+2))/(3+1-4)

    Hmmm. STill 0/0. I guess we have to try l'Hospital's Rule. Taking derivatives top and bottom we get

    1/√(1+2x) - √3/(2√x)
    -------------------------
    1/(2√(3+x) - 1/√x

    -> (1/√3 - √3/2)/(1/2√4 - 1/√1)
    = 2/(3√3)

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