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Math Help - Simple limit with radical, but I keep getting the wrong answer.

  1. #1
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    Simple limit with radical, but I keep getting the wrong answer.

    Hello. I've worked this problem several times, but I keep coming up with the wrong answer and it's driving me mad. Here it is:



    The answer is 1/6, but I keep getting undefined. I multiply by the conjugate, then reduce, and I get: \frac{1}{\sqrt{x+7}-3} I have no clue where I'm messing up with the algebra.

    Thanks for any help.
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  2. #2
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    \frac{{\sqrt {x + 7}  - 3}}<br />
{{x - 2}} = \frac{{(x + 7) - 9}}<br />
{{(x - 2)\left( {\sqrt {x + 7}  + 3} \right)}}
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  3. #3
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    Hello, Sirmio!

    Your algebra must be off . . .


    \lim_{x\to2}\,\frac{\sqrt{x+7} - 3}{x-2}

    Multiply top and bottom by the conjugate:

    . . \frac{\sqrt{x+7}-3}{x-2}\cdot\frac{\sqrt{x+7}+3}{\sqrt{x+7}+3} \;=\;\frac{(x+7) - 9}{(x-2)(\sqrt{x+7} + 3)} . =\;\frac{{\color{red}\rlap{/////}}x-2}{({\color{red}\rlap{/////}}x-2)(\sqrt{x+7}+3)} \;=\;\frac{1}{\sqrt{x+7}+3}


    Therefore: . \lim_{x\to2}\frac{1}{\sqrt{x+7}+3} \;=\;\frac{1}{\sqrt{9}+3} \;=\;\frac{1}{6}

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  4. #4
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    Lightbulb

    Thanks! I didn't change the sign on the conjugate. I knew it was simple. Thanks for the help!
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