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Math Help - Limits

  1. #1
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    Limits

    how do i find the value of lim of (2x-6) over (square of x - square root of 3) as x approaches 3 if limit exist.
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  2. #2
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    \frac{2x-6}{\sqrt{x}-\sqrt{3}}=\frac{2\left( \sqrt{x}+\sqrt{3} \right)\left( \sqrt{x}-\sqrt{3} \right)}{\sqrt{x}-\sqrt{3}}=2\left( \sqrt{x}+\sqrt{3} \right)\to 4\sqrt{3} as x\to3.
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  3. #3
    Eater of Worlds
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    This this what you mean?:

    \lim_{x\to 3}\frac{2x-6}{(x-\sqrt{3})^{2}}

    or is it:

    \lim_{x\to 3}\frac{2x-6}{x^{2}-\sqrt{3}}

    or none of the above?.
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  4. #4
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    Quote Originally Posted by galactus View Post
    This this what you mean?:

    \lim_{x\to 3}\frac{2x-6}{(x-\sqrt{3})^{2}}

    or is it:

    \lim_{x\to 3}\frac{2x-6}{x^{2}-\sqrt{3}}

    or none of the above?.

    like the bottom one except that its not x^2 its square root of x
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  5. #5
    Eater of Worlds
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    You had 'square of x', not square root of x. Anyway, Kriz nailed it.
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