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Math Help - One Sided Limit Problem

  1. #1
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    One Sided Limit Problem

    I need to find the lim as x approaches 9 from the left of this function:

    f(x) = ((sqrt(x))-3)/(x-9)

    I know from creating a table of y values when x is near 9 and from looking at its graph that the limit from either side is about 1.6666... , but I have been trying to manipulate the function to get in into a form that allows me to put 9 into the equation and actually solve for an exact answer. I can seem to figure out how. Any ideas?
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  2. #2
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    Re: One Sided Limit Problem

    Hey mattjp213.

    You might want to check if you can use l'Hopitals rule here given the nature of your limit (it should).
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  3. #3
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    Re: One Sided Limit Problem

    Hint:  (\sqrt{x}+3) \text{ is a factor of } (x-9) and so is (...)

    You'll get  \lim_{x \to 9}f(x) = \frac{1}{6} then you're done.
    Last edited by chen09; September 25th 2013 at 08:03 PM.
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  4. #4
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    Re: One Sided Limit Problem

    Hello, mattjp213!

    \displaystyle \lim_{x\to9^-}\frac{\sqrt{x}-3}{x-9}

    Multiply by \frac{\sqrt{x} + 3}{\sqrt{x}-3}

    . . \lim_{x\to9^-}\frac{\sqrt{x}-3}{x-9}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+3} \;=\; \lim_{x\to9^-}\frac{x-9}{(x-9)(\sqrt{x}+3)}

    . . . =\; \lim_{x\to9^-}\frac{1}{\sqrt{x} + 3} \;=\;\frac{1}{\sqrt{9}+3} \;=\;\frac{1}{3+3} \;=\;\frac{1}{6}

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  5. #5
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    Re: One Sided Limit Problem

    Quote Originally Posted by Soroban View Post
    Hello, mattjp213!


    Multiply by \frac{\sqrt{x} + 3}{\sqrt{x}-3}

    . . \lim_{x\to9^-}\frac{\sqrt{x}-3}{x-9}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+3} \;=\; \lim_{x\to9^-}\frac{x-9}{(x-9)(\sqrt{x}+3)}

    . . . =\; \lim_{x\to9^-}\frac{1}{\sqrt{x} + 3} \;=\;\frac{1}{\sqrt{9}+3} \;=\;\frac{1}{3+3} \;=\;\frac{1}{6}

    Thanks for giving him the exact solution so he doesn't have to think about it.
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