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Math Help - Calculating a "0/0" limit

  1. #1
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    Calculating a "0/0" limit

    How can I calculate the following limit without using l'H˘pital's rule?

    \lim_{x\to 2}\frac {\sqrt{x+2}-\sqrt{3x-2}}{\sqrt{4x+1}-\sqrt{5x-1}}
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  2. #2
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    Re: Calculating a "0/0" limit

    Quote Originally Posted by loui1410 View Post
    How can I calculate the following limit without using l'H˘pital's rule?
    \lim_{x\to 2}\frac {\sqrt{x+2}-\sqrt{3x-2}}{\sqrt{4x+1}-\sqrt{5x-1}}
    The original expression is equal \left( {\frac{{ - 2x + 4}}{{ - x + 2}}} \right)\left( {\frac{{\sqrt {4x + 1}  + \sqrt {5x - 1} }}{{\sqrt {x + 2}  + \sqrt {3x - 2} }}} \right).
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  3. #3
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    Re: Calculating a "0/0" limit

    I got to that and didn't know how to continue.. Only now did I notice that we can simply reduce the fraction Thanks.
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    Re: Calculating a "0/0" limit

    Quote Originally Posted by loui1410 View Post
    I got to that and didn't know how to continue.. Only now did I notice that we can simply reduce the fraction Thanks.
    \left( {\frac{{ - 2x + 4}}{{ - x + 2}}} \right)\left( {\frac{{\sqrt {4x + 1} + \sqrt {5x - 1} }}{{\sqrt {x + 2} + \sqrt {3x - 2} }}} \right)\to(2)\left( {\frac{6}{4} \right)
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