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Math Help - A limit that finally stumped me!

  1. #1
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    A limit that finally stumped me!

    So anyone have any idea on how to solve this one? I've tried all the usual ways to solve..getting no where fast.

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  2. #2
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    Hello,

    The conjugate of a number a-b is a+b (and conversely). (a-b)(a+b)=a-b. This is why it can be very useful when you want to get rid of some square roots, or transforming a difference into a sum.

    Multiply by 1=\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}

    Last edited by Moo; September 10th 2008 at 01:55 PM.
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    Ok - so I multiplied by the conjugate, as normal and I got:


    <br />
\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}<br />

    So I still zero out the denom as I come to 2...help here?
    Last edited by 66replica; September 10th 2008 at 02:03 PM.
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  4. #4
    Moo
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    Quote Originally Posted by 66replica View Post
    Ok - so I multiplied by the conjugate, as normal and I got:


    <br />
\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{2-x}<br />

    So I still zero out the denom as I come to 2...help here?
    No, pardon me ! I misread and actually it was \sqrt{3-x}+1 for the conjugate !

    Now, expand (\sqrt{6-x}-2)(\sqrt{3-x}+1). In general, it must be a factor of x-2
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,

    The conjugate of a number a-b is a+b (and conversely). (a-b)(a+b)=a-b. This is why it can be very useful when you want to get rid of some square roots, or transforming a difference into a sum.

    Multiply by 1=\frac{\sqrt{3-x}-1}{\sqrt{3-x}-1}

    Small mistake:
    Multiply by 1=\frac{\sqrt{3-x}{\color{red}+}1}{\sqrt{3-x}{\color{red}+}1}

    To the OP: I'm very surprised that what moo showed you wasn't on your check list of all the usual ways ..... Clearly moo made a typo that you should have realised and corrected for.
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  6. #6
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    Yeah it was a typo on my part. I've done the conjugate thing a few times on this problem and keep getting stuck trying to get an x-2 factored from the top.

    I get stuck even after I foil the numerator I end up with:

    <br />
\frac{(\sqrt{6-x})(\sqrt{3-x})-2(\sqrt{3-x}+\sqrt{6-x}-2}{2-x}<br />
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  7. #7
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    Quote Originally Posted by 66replica View Post
    Yeah it was a typo on my part. I've done the conjugate thing a few times on this problem and keep getting stuck trying to get an x-2 factored from the top.

    I get stuck even after I foil the numerator I end up with:

    <br />
\frac{(\sqrt{6-x})(\sqrt{3-x})-2(\sqrt{3-x}+\sqrt{6-x}-2}{2-x}<br />
    You get \frac{(\sqrt{6 - x} - 2)(\sqrt{3 - x} + 1)}{2-x}.

    Now multiply this expression by \frac{(\sqrt{6 - x} + 2)}{\sqrt{6 - x} + 2}.

    You get \frac{(2 - x) (\sqrt{3 - x} + 1)}{(2-x) (\sqrt{6 - x} + 2)} \, ....
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  8. #8
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    Thanks for the help - I understand the problem well now.
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  9. #9
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    Wouldn't it be easier just to use L'hopital's rule? Then you would get \frac{\sqrt{3-x}}{\sqrt{6-x}} | x=2 \Rightarrow 1/2
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  10. #10
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    Quote Originally Posted by Hoover glasses View Post
    Wouldn't it be easier just to use L'hopital's rule? Then you would get \frac{\sqrt{3-x}}{\sqrt{6-x}} | x=2 \Rightarrow 1/2
    Yes but most people don't learn L'hopital's rule until later in their calc study
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