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Math Help - Finding a limit involving two different radicals

  1. #1
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    Finding a limit involving two different radicals

    \lim_{x\rightarrow4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5}

    Every attempt at multiplying either the top or bottom by the negated counterpart ends up leading to a 0/0 answer. I know there's some trick I'm missing to solve this, could some one point me in the right direction without giving away the answer?
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  2. #2
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    Quote Originally Posted by penguinpwn View Post
    \lim_{x\rightarrow4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5}

    Every attempt at multiplying either the top or bottom by the negated counterpart ends up leading to a 0/0 answer. I know there's some trick I'm missing to solve this, could some one point me in the right direction without giving away the answer?
    \displaystyle \lim_{x\to4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5} \cdot \frac{\sqrt{8-x}+2}{\sqrt{8-x}+2} \cdot \frac{\sqrt{29-x}+5}{\sqrt{29-x}+5}

    \displaystyle \lim_{x\to4} \frac{[(8-x)-4] \cdot [\sqrt{29-x}+5]}{[\sqrt{8-x}+2] \cdot [(29-x)-25]}

    finish it
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \displaystyle \lim_{x\to4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5} \cdot \frac{\sqrt{8-x}+2}{\sqrt{8-x}+2} \cdot \frac{\sqrt{29-x}+5}{\sqrt{29-x}+5}

    \displaystyle \lim_{x\to4} \frac{[(8-x)-4] \cdot [\sqrt{29-x}+5]}{[\sqrt{8-x}+2] \cdot [(29-x)-25]}

    finish it
    Gahhh my mind isn't abstract enough for calculus...

    The answer is 5/2

    Thank you Skeeter.
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