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Math Help - I'm not sure if I got this one right

  1. #1
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    I'm not sure if I got this one right

    <br />
\lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}

    and what I got was

    <br />
\frac{1}{\sqrt{5}+\sqrt{5}}<br />
    Last edited by drkidd22; July 28th 2009 at 06:40 AM.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by drkidd22 View Post
    <br />
\lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}
    \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 5}  - \sqrt 5 }}<br />
{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {x + 5}  - \sqrt 5 } \right)\left( {\sqrt {x + 5}  + \sqrt 5 } \right)}}<br />
{{x\left( {\sqrt {x + 5}  + \sqrt 5 } \right)}} =

    = \mathop {\lim }\limits_{x \to 0} \frac{{x + 5 - 5}}<br />
{{x\left( {\sqrt {x + 5}  + \sqrt 5 } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {x + 5}  + \sqrt 5 }} =

    = \frac{1}{{\sqrt {0 + 5}  + \sqrt 5 }} = \frac{1}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{{10}}.
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  3. #3
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    Oh ok, so I did do it right, just didn't complete the whole thing.
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  4. #4
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    Quote Originally Posted by drkidd22 View Post
    <br />
\lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}

    and what I got was

    <br />
\frac{1}{\sqrt{5}+\sqrt{5}}<br />
    In the future, when you ask for a check on what you did, it would be better to show how you got the result rather than just the result.
    Rationalize the numerator by multiplying numerator and denominator by \sqrt{x+5}+ \sqrt{5}:
    \frac{x+ 5- x}{x(\sqrt{x+5}+ \sqrt{x})}= \frac{x}{x(\sqrt{x+5}+ \sqrt{5}} \frac{1}{\sqrt{x+5}+ \sqrt{5}}.
    Now you can put x= 0 and get \frac{1}{\sqrt{5}+\sqrt{5}}, what you have. But I would recommend that you write as \frac{1}{2\sqrt{5}} or even \frac{\sqrt{5}}{10}.

    Blast! DeMath got in ahead of me!
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  5. #5
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    This is anotherone I tried doing today.

    <br /> <br />
\lim_{x\to 0} = \frac{\frac{1}{3+x}-\frac{1}{3}}{x}<br /> <br />

    <br /> <br />
\frac{\frac{3-3+x}{3(3+x)}}{x}<br /> <br /> <br />

    <br />
\frac{\frac{x}{9+3x}}{x}<br />
=9?<br />
    Last edited by drkidd22; July 28th 2009 at 07:54 AM.
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  6. #6
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    Quote Originally Posted by drkidd22 View Post
    This is anotherone I tried doing today.

    <br /> <br />
\lim_{x\to 0} = \frac{\frac{1}{3+x}-\frac{1}{3}}{x}<br /> <br />

    <br /> <br />
\frac{\frac{3-3+x}{3(3+x)}}{x}<br /> <br /> <br />

    <br />
\frac{\frac{x}{9+3x}}{x}<br />
=9?<br />
    \frac{\frac{3-(3+x)}{3(3+x)}}{x} = \frac{-x}{3x(3+x)} = \frac{-1}{3(3+x)}

    limit is -\frac{1}{9}

    in future, start a new problem with a new thread.
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