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Math Help - How would you approach this

  1. #1
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    How would you approach this

    Find the values for the constants a and b such that
    \lim[x->o]{\frac{\sqrt{a+bx}-\sqrt{3}}{x}}=\sqrt{3}

    As the title suggests, I'm stuck as to how one would approach this problem.
    I can remove the roots, but I don't know what I should do to remove the x from the denominator. I'm thinking that b = all real, a = 12 but I can't prove it.
    Thoughts? Suggestions?
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  2. #2
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    Quote Originally Posted by Guess992 View Post
    Find the values for the constants a and b such that
    \lim[x->o]{\frac{\sqrt{a+bx}-\sqrt{3}}{x}}=\sqrt{3}

    As the title suggests, I'm stuck as to how one would approach this problem.
    I can remove the roots, but I don't know what I should do to remove the x from the denominator. I'm thinking that b = all real, a = 12 but I can't prove it.
    Thoughts? Suggestions?
    a = 3 , because any other value will make the limit nonexistent.


    \lim_{x \to 0} \frac{\sqrt{3+bx}-\sqrt{3}}{x} \cdot \frac{\sqrt{3+bx} + \sqrt{3}}{\sqrt{3+bx} + \sqrt{3}} =\sqrt{3}

    \lim_{x \to 0} \frac{(3+bx)-3}{x(\sqrt{3+bx} + \sqrt{3})} = \sqrt{3}

    \lim_{x \to 0} \frac{bx}{x(\sqrt{3+bx} + \sqrt{3})} = \sqrt{3}

    \lim_{x \to 0} \frac{b}{\sqrt{3+bx} + \sqrt{3}} = \sqrt{3}

    \frac{b}{2\sqrt{3}} = \sqrt{3}

    b = 6
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  3. #3
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    Could you explain how it fails to exist if a =/= 3? I don't quite follow how you get to that point.
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  4. #4
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    Quote Originally Posted by Guess992 View Post
    Could you explain how it fails to exist if a =/= 3? I don't quite follow how you get to that point.
    let a = any value except 3 and determine the limit as x \to 0 ... see what happens.
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  5. #5
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    As x goes to 0, the numerator goes to 0, the numerator goes to \sqrt{a}- \sqrt{3} while the denominator goes to 0. If the numerator does NOT go to 0, then the fraction gets larger and larger without bound. There would be no limit.
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