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Math Help - limits help

  1. #1
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    limits help

    problem:

    lim as x -> 4 : 4-x / (5 - sqrt(x^2 + 9))

    I tried to use the conjugate property but in the end when I substitute 4 into the equation I come up with it turns to:

    (4 - 4) * (5 + sqrt(4^2 + 9)) / 25 - 16 + 9
    0 * 5 + 5 / 18

    the answer should be 5/4, where am I going wrong with this problem?
    Last edited by deltemis; February 4th 2014 at 05:25 PM.
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  2. #2
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    Re: limits help

    Quote Originally Posted by deltemis View Post
    problem:

    lim as x -> 4 : 4-x / 5 - sqrt(x^2 + 9)

    I tried to use the conjugate property but in the end when I substitute 4 into the equation I come up with it turns to:

    (4 - 4) * (5 + sqrt(4^2 + 9)) / 25 - 16 + 9
    0 * 5 + 5 / 18

    the answer should be 5/4, where am I going wrong with this problem?
    we need to get the parens squared away first. Is this what you mean?

    \displaystyle{\lim_{x \to 4}}\frac{4-x}{5}-\sqrt{x^2+9}

    or do you maybe mean this?

    \displaystyle{\lim_{x \to 4}}\frac{4-x}{5-\sqrt{x^2+9}}

    I think you mean the 2nd one.

    \frac{4-x}{5-\sqrt{x^2+9}}=\frac{4-x}{5-\sqrt{x^2+9}}\frac{5+\sqrt{x^2+9}}{5+\sqrt{x^2+9}}  =

    \frac{(4-x)\left(5+\sqrt{x^2+9}\right)}{5^2-(x^2+9)}=\frac{(4-x)\left(5+\sqrt{x^2+9}\right)}{16-x^2}=

    \frac{(4-x)\left(5+\sqrt{x^2+9}\right)}{(4-x)(4+x)}=\frac{5+\sqrt{x^2+9}}{4+x}

    Now you can just plug 4 in for x to find the limit

    \displaystyle{\lim_{x \to 4}}\frac{4-x}{5+\sqrt{x^2+9}}=\frac{5+\sqrt{4^2+9}}{(4+4)}=

    \frac{10}{8}= \frac{5}{4}
    Last edited by romsek; February 4th 2014 at 05:34 PM.
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  3. #3
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    Re: limits help

    it's what you have but the sqrt portion is part of the denominator
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  4. #4
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    Re: limits help

    Quote Originally Posted by deltemis View Post
    problem:

    lim as x -> 4 : 4-x / 5 - sqrt(x^2 + 9)

    I tried to use the conjugate property but in the end when I substitute 4 into the equation I come up with it turns to:
    (4 - 4) * (5 + sqrt(4^2 + 9)) / 25 - 16 + 9
    0 * 5 + 5 / 18

    the answer should be 5/4, where am I going wrong with this problem?
    I think that you mean something else altogether.
    \frac{4-x}{5-\sqrt{x^2+9}}=\frac{(4-x)(5+\sqrt{x^2+9})}{25-(x^2+9)}=\frac{5+\sqrt{x^2+9}}{4+x}
    Last edited by Plato; February 4th 2014 at 05:34 PM.
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