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Math Help - Simple Limit problem

  1. #1
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    Simple Limit problem

    \frac{49-x}{\sqrt{x}-7}
    [tex]\lim_{x \to 49}[/Math]


    \frac{49-x}{\sqrt{x}-7}*\frac{\sqrt{x}+7}{\sqrt{x+7}}=\frac{49\sqrt{x}+  343-x\sqrt{x}}{x-49}

    not sure where to go from here. I tried factoring out (x-49), but I still end up with an undefined answer.

    What am I missing? (the answer is not undefined!)

    Thanks!
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  2. #2
    MHF Contributor harish21's Avatar
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     49 - x = (7)^2 - (\sqrt{x})^2 = (7 - \sqrt{x})(7+\sqrt{x})= -(\sqrt{x}-7)(7+\sqrt{x})

    or

    use Lhopitals rule to get the derivatives of the numerator and the denominator.
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  3. #3
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    Quote Originally Posted by Vamz View Post
    \frac{49-x}{\sqrt{x}-7}
    [tex]\lim_{x \to 49}[/Math]


    \frac{49-x}{\sqrt{x}-7}*\frac{\sqrt{x}+7}{\sqrt{x+7}}=\frac{49\sqrt{x}+  343-x\sqrt{x}}{x-49}

    not sure where to go from here. I tried factoring out (x-49), but I still end up with an undefined answer.

    What am I missing? (the answer is not undefined!)

    Thanks!
    Your numerator is incomplete...

    (49-x)(\sqrt{x}+7)=49(\sqrt{x}+7)-x(\sqrt{x}+7)=49\sqrt{x}+49(7)-x\sqrt{x}-7x

    If x is 49, the numerator is 49(7)+49(7)-49(7)-7(49)

    which means we've gotten nowhere!
    Both numerator and denominator are still zero if x=49.

    This is because we are still subtracting equal quantities in both numerator and denominator.

    Notice how Harish's method leaves us without subtractions.
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  4. #4
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    Quote Originally Posted by Vamz View Post
    \frac{49-x}{\sqrt{x}-7}
    \lim_{x \to 49}


    \frac{49-x}{\sqrt{x}-7}*\frac{\sqrt{x}+7}{\sqrt{x+7}}=\frac{49\sqrt{x}+  343-x\sqrt{x}}{x-49}

    not sure where to go from here. I tried factoring out (x-49), but I still end up with an undefined answer.

    What am I missing? (the answer is not undefined!)

    Thanks!
    Substitute x = t^2 and your limit becomes \displaystyle \lim_{t \to 7} \frac{49-t^2}{t-7}. The solution should be obvious.
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