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Math Help - Compute the limit

  1. #1
    Member rowe's Avatar
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    Compute the limit

    Compute \lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}.

    I approached this by multiplying the whole expression by the conjugate of the numerator, \sqrt{6-x}+2, which gave me a non-zero denominator, but evaluated the limit as 0. The actual limit is \tfrac{1}{2}.

    What's the trick with this one?
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  2. #2
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    \frac{{\sqrt {6 - x}  - 2}}{{\sqrt {3 - x}  - 1}} = \frac{{(2 - x)\left( {\sqrt {3 - x}  + 1} \right)}}{{(2 - x)\left( {\sqrt {6 - x}  + 2} \right)}}
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  3. #3
    Member rowe's Avatar
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    Quote Originally Posted by Plato View Post
    \frac{{\sqrt {6 - x}  - 2}}{{\sqrt {3 - x}  - 1}} = \frac{{(2 - x)\left( {\sqrt {3 - x}  + 1} \right)}}{{(2 - x)\left( {\sqrt {6 - x}  + 2} \right)}}
    Thanks. Could you perhaps show me the steps you took to achieve that factorisation? Either numerator or denominator will do, I think
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  4. #4
    Math Engineering Student
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    2-x=\left( \sqrt{6-x}+2 \right)\left( \sqrt{6-x}-2 \right)=\left( \sqrt{3-x}+1 \right)\left( \sqrt{3-x}-1 \right).
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  5. #5
    Junior Member BugzLooney's Avatar
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    How are you guys getting all these fancy symbols? Do I not have enough posts yet or something?
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  6. #6
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    It's using an inbuilt LaTeX compiler on this site.

    For tutorials, see the LaTeX subforum.
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