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Thread: 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

    $\displaystyle
    \lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}$

    and what I got was

    $\displaystyle
    \frac{1}{\sqrt{5}+\sqrt{5}}
    $
    Last edited by drkidd22; Jul 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
    $\displaystyle
    \lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}$
    $\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 5} - \sqrt 5 }}
    {x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {x + 5} - \sqrt 5 } \right)\left( {\sqrt {x + 5} + \sqrt 5 } \right)}}
    {{x\left( {\sqrt {x + 5} + \sqrt 5 } \right)}} =$

    $\displaystyle = \mathop {\lim }\limits_{x \to 0} \frac{{x + 5 - 5}}
    {{x\left( {\sqrt {x + 5} + \sqrt 5 } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {x + 5} + \sqrt 5 }} =$

    $\displaystyle = \frac{1}{{\sqrt {0 + 5} + \sqrt 5 }} = \frac{1}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{{10}}.$
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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
    $\displaystyle
    \lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}$

    and what I got was

    $\displaystyle
    \frac{1}{\sqrt{5}+\sqrt{5}}
    $
    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 $\displaystyle \sqrt{x+5}+ \sqrt{5}$:
    $\displaystyle \frac{x+ 5- x}{x(\sqrt{x+5}+ \sqrt{x})}= \frac{x}{x(\sqrt{x+5}+ \sqrt{5}}$$\displaystyle \frac{1}{\sqrt{x+5}+ \sqrt{5}}$.
    Now you can put x= 0 and get $\displaystyle \frac{1}{\sqrt{5}+\sqrt{5}}$, what you have. But I would recommend that you write as $\displaystyle \frac{1}{2\sqrt{5}}$ or even $\displaystyle \frac{\sqrt{5}}{10}$.

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

    $\displaystyle

    \lim_{x\to 0} = \frac{\frac{1}{3+x}-\frac{1}{3}}{x}

    $

    $\displaystyle

    \frac{\frac{3-3+x}{3(3+x)}}{x}


    $

    $\displaystyle
    \frac{\frac{x}{9+3x}}{x}
    =9?
    $
    Last edited by drkidd22; Jul 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.

    $\displaystyle

    \lim_{x\to 0} = \frac{\frac{1}{3+x}-\frac{1}{3}}{x}

    $

    $\displaystyle

    \frac{\frac{3-3+x}{3(3+x)}}{x}


    $

    $\displaystyle
    \frac{\frac{x}{9+3x}}{x}
    =9?
    $
    $\displaystyle \frac{\frac{3-(3+x)}{3(3+x)}}{x} = \frac{-x}{3x(3+x)} = \frac{-1}{3(3+x)}$

    limit is $\displaystyle -\frac{1}{9}$

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