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Math Help - Evaluating limit

  1. #1
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    Evaluating limit

    Not sure how to go about this.

    Can I just rationalize either the numerator or denominator?

    lim (√2-x) - (√8-4x)
    x->2 (√3x+3) - (√11-x)
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  2. #2
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    Quote Originally Posted by Neconine View Post
    Not sure how to go about this.

    Can I just rationalize either the numerator or denominator?

    lim (√2-x) - (√8-4x)
    x->2 (√3x+3) - (√11-x)
    Hi

    The numerator is \sqrt{2-x}-\sqrt{4(2-x)} = \sqrt{2-x}-2\sqrt{2-x} = -\sqrt{2-x}


    \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}}\:\:\frac{\sqrt{3x+3}+\sqrt{11-x}}{\sqrt{3x+3}+\sqrt{11-x}}


    \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}\left(\sqrt{3x+3}+\sqrt{11-x}\right)}{4x-8}


    \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} =  \frac{-\sqrt{2-x}\left(\sqrt{3x+3}+\sqrt{11-x}\right)}{-4(2-x)} =  \frac{\sqrt{3x+3}+\sqrt{11-x}}{4\sqrt{2-x}}
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  3. #3
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    Quote Originally Posted by Neconine View Post
    Not sure how to go about this.

    Can I just rationalize either the numerator or denominator?

    lim (√2-x) - (√8-4x)
    x->2 (√3x+3) - (√11-x)
    first, note that this limit should be one-sided since the domain of the rational expression is -1 \le x < 2

    \lim_{x \to 2^-} \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}}

    \lim_{x \to 2^-}  \frac{\sqrt{2-x}-2\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}}

    \lim_{x \to 2^-}  \frac{-\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}} \cdot \frac{\sqrt{3x+3}+\sqrt{11-x}}{\sqrt{3x+3}+\sqrt{11-x}}

    \lim_{x \to 2}  \frac{-\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{ (3x+3)-(11-x)}

    \lim_{x \to 2^-}  \frac{-\sqrt{2-x}  \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{4x-8}

    \lim_{x \to 2}  \frac{\sqrt{2-x}  \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{8-4x}

    \lim_{x \to 2^-}  \frac{\sqrt{2-x}  \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{4(2-x)}

    \lim_{x \to 2^-}  \frac{\sqrt{3x+3}+\sqrt{11-x}}{4\sqrt{2-x}}

    as x \to 2^- , the numerator approaches 6 and the denominator approaches 0 ... the limit does not exist.
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