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Math Help - Limit question, likely a simple mistake on my part.

  1. #1
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    Limit question, likely a simple mistake on my part.

    Hey

    Question:
    Evaluate the following limit: \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x}

    Solution:
    I tried multiplying by the conjugate, but that gives me:  \frac {x - 9}{(9 - x)(\sqrt {x} + 3)}

    and I really don't know what to do from here. The top now gives 0, and I can't factor it out using the bottom. Help?
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Kakariki View Post
    Hey

    Question:
    Evaluate the following limit: \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x}

    Solution:
    I tried multiplying by the conjugate, but that gives me:  \frac {x - 9}{(9 - x)(\sqrt {x} + 3)}

    and I really don't know what to do from here. The top now gives 0, and I can't factor it out using the bottom. Help?
    Your method is correct!  \frac {x - 9}{(9 - x)(\sqrt {x} + 3)} = \frac {x - 9}{(-(x-9))(\sqrt {x} + 3)} = \frac{1}{-(\sqrt {x} + 3)} = \frac{-1}{6}

    OR,

    dont multiply. factorize the denominator:

    \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(x-9)} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(\sqrt{x} - 3)(\sqrt{x} + 3)}=  \lim_{x\to9} \frac{1}{\sqrt{x} + 3} = \frac{-1}{6}
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  3. #3
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    Quote Originally Posted by harish21 View Post
    Your method is correct!  \frac {x - 9}{(9 - x)(\sqrt {x} + 3)} = \frac {x - 9}{(-(x-9))(\sqrt {x} + 3)} = \frac{1}{-(\sqrt {x} + 3)} = \frac{-1}{6}

    OR,

    dont multiply. factorize the denominator:

    \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(x-9)} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(\sqrt{x} - 3)(\sqrt{x} + 3)}=  \lim_{x\to9} \frac{1}{\sqrt{x} + 3} = \frac{-1}{6}
    Oh dear, that is an easy thing to do. I just factored the negative out of the 9 - x and then got - 1/6. that's really easy
    Thanks!!!
    BTW, if that is Iron maiden as your avatar, you are awesome. If not, you are still awesome for helping me
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Kakariki View Post
    Oh dear, that is an easy thing to do. I just factored the negative out of the 9 - x and then got - 1/6. that's really easy
    Thanks!!!
    BTW, if that is Iron maiden as your avatar, you are awesome. If not, you are still awesome for helping me
    Haha! Yes, Thats Iron Maiden.
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